Taylor Series Problem

  • #1
I have got a question here that puzzles me.

How do I use TAYLOR SERIES to find the 2005th derivative for the function when x=0 for the following function:

f(x) = inverse tan [(1+x)/(1-x)]

Part (1) I was hinted that differentiating inverse tan x is = 1/(1+x^2).

Part (2) After which, I need to integrate 1/(1+x^2) and include some ARBITRARY constants to get back inverse tan [(1+x)/(1-x)]

I would like to know how whats the approach for part (2) as I am very confuse now.

Hope you can help. Thanks! :shy:
 

Answers and Replies

  • #2
Hurkyl
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I don't know why you'd be confused, so let's take a step back. What Taylor series did you get as the answer to part (1)?
 
  • #3
Well. the series for the part (1) is a geometric series
I get Summation from n=0 to infinity : (-1)^n * (x)^2n

But i just couldn't see/figure out what part (2) is saying. Including some arbitrary constant? Pardon me if I am abit slow. I just couldn't see the link here.
How do I determine the arbitrary constant?
 
  • #4
quasar987
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chuachinghong said:
How do I use TAYLOR SERIES to find the 2005th derivative
Haha! This reminds me of my first "multivariable calc" professor who was exasperated at us during the correction of our exam from realizing that half the class had manually computed 12 derivatives of f to answer the question "knowing the maclaurin series of f is ..., compute the 12th derivative of f at x=0." :rofl:

Maybe your teacher had a similar experience int he past years! :rofl:
 
  • #5
shmoe
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chuachinghong said:
How do I determine the arbitrary constant?

By evaluating your series and f(x) at the same convenient value of x. I'd suggest x=0 is very convenient.

You should first ask though, does this arbitrary constant matter at all when looking at the 2005th derivative?
 
Last edited:
  • #6
Hello shmoe, what I done was to get the TAYLOR SERIES for that
inverse tan [(1+x)/(1-x)]. I got that by subsituting [(1+x)/(1-x)] into the X of the TAYLOR SERIES of inverse tan X.
Then I compare it with the TAYLOR SERIES definition/formula to find the 2005th derivative when x=0.

Just want to know whether my approach is correct or not?
 
Last edited:
  • #7
shmoe
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chuachinghong said:
Hello shmoe, what I done was to get the TAYLOR SERIES for that
inverse tan [(1+x)/(1-x)]. I got that by subsituting [(1+x)/(1-x)] into the X of the TAYLOR SERIES of inverse tan X.

This seems like a disaster will happen. Try finding the Taylor series of inverse tan [(1+x)/(1-x)] directly, what is the derivative of inverse tan [(1+x)/(1-x)]?

ps. You don't have to capitalize all of "Taylor series", just the "T" will do.
 
  • #8
Thanks everyone. I got it now. @,@
 

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