Maclaurin series for integrals

  • #1

Homework Statement


Heres the question: write the first three nonzero terms (Maclaurin Series)

[tex]\int^{1+sinx}_{1}(1/(sqrt{(1+ln(x))})dx[/tex]



Homework Equations





The Attempt at a Solution


so for the similar questions, i use maclaurin series for common functions (you can see it from this website http://mathworld.wolfram.com/MaclaurinSeries.html" [Broken])
for this question, i tried to start with lnx, but from what i searched online, there is no maclaurin series for lnx (because it is not defined at 0), so now i am stuck. also, i tried to inegrate just the function itself, but it seems impossible.
if lnx does not have a maclaurin series, that function does not have a maclaurin series?
 
Last edited by a moderator:

Answers and Replies

  • #2
lanedance
Homework Helper
3,304
2
Hi bobthetomato - welcome to PF ;)

I think the notation is a little confusing, does it help if you re-write in terms of a dummy variable?
[tex]f(x) = \int^{1+sinx}_{1} \frac{du}{\sqrt{1+ln(u)}}[/tex]

note the intergal is over u from 1 to 1 + sin(x), so the ln(u) doesn't blow up

then you can differntiate w.r.t. x to find each term, which could be a little tricky...
 
  • #3
thank your lanedance, that function looks better now.
i dont understand, why do i need to differentiate it? can i integrate the whole thing? but if i integrate the equation, i wont get maclaurin series (first three terms).

this is what i tried:

since there is a series for ln(1-x) = -x-(x^2/2)-...

let x=1-t

ln(1-(1-t))=-(1-t)-((1-t)^2/2)-...

ln(t)=-(1-t)-((1-t)^2/2)-...

but if i continue to do it, it becomes complicated and hard to differentiate

1+ln(t)= t-((1-t)^2/2)-...

1/(1+ln(t))^(1/2)= 1/(t-((1-t)^2/2)-...)^(1/2)

i stopped here because i do not know how to integrate this function.
 
  • #4
Can someone please help me? This is only a university first year math question, so I think it shouldn't be that hard.

Here is the question again...
[tex]
f(x) = \int^{1+sinx}_{1} \frac{dt}{\sqrt{1+ln(t)}}
[/tex]

so i think substitution is the right place to start.

as the range of integration is near 1 (because 1+sinx can range from 1 to 2), ln(t) will not blow up.

so i set x=(1-t)

[tex]\ln(1-x) = -x - \frac{x^{2}}{2} - \frac{x^{3}}{3}+...[/tex]

if i substitute 1-t into all the x

[tex]\ln(1-(1-t)) = -(1-t) - \frac{(1-t)^{2}}{2} - \frac{(1-t)^{3}}{3}+...[/tex]

and then it turns out like this

[tex]\ln(t) = -1+t - \frac{(1-t)^{2}}{2} - \frac{(1-t)^{3}}{3}+...[/tex]

now if I add 1 to the whole equation

[tex]\ 1 + ln(t) = 1 -1+t - \frac{(1-t)^{2}}{2} - \frac{(1-t)^{3}}{3}+...[/tex]

which will equal to

[tex]\ 1 + ln(t) = t - \frac{(1-t)^{2}}{2} - \frac{(1-t)^{3}}{3}+...[/tex]

and now if i square root the whole thing and 1/the whole equation, i will get the this...
oh and i added integrals

[tex]
f(x) = \int^{1+sinx}_{1} \frac{dt}{\sqrt{t-\frac{(1-t)^{2}}{2}-\frac{(1-t)^{3}}{3}+...}}
[/tex]

so know i have made it to Maclaurin Series like, i do not know what to do next. How do you simplify and integrate this?
This question is bugging me, please help!
 
  • #5
lanedance
Homework Helper
3,304
2
sorry if is miss something, tex doesn't display well on this computer..
so you need to evaluate the macluarin sereis of:
[tex]f(x) = \int^{1+sinx}_{1} \frac{dt}{\sqrt{1+ln(t)}}[/tex]

so rather than substituting, expanding etc. why not just evaulate the derivatives
f'(x), f''(x) and evaluate at x = 0. f(0) should be obvious

As an example, consider:
[tex] F(x) = \int_0^x du u^2 = \frac{u^3}{3}|_0^x = \frac{x^3}{3}[/tex]
[tex] F'(x) = f(x) = \frac{d}{dx}\int_0^x du u^2 = \frac{d}{dx}\frac{x^3}{3}=x^2[/tex]
[tex] F''(x) = f'(x) = ...[/tex]
now all these should be easy to evaluate at x=0, how can you write the maclaurin expansion using those terms? (though its a very trivial expnasion for a single polynomial, but its the method that is important here)
 
  • #6
lanedance
Homework Helper
3,304
2
PS if its still not clear have a look at the definition of maclaurin series in terms of derivatives of the function
 

Related Threads on Maclaurin series for integrals

  • Last Post
Replies
4
Views
7K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
2
Views
2K
Replies
11
Views
7K
  • Last Post
Replies
3
Views
943
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
613
  • Last Post
Replies
8
Views
719
Top