Where does this method of constructing the taylor expansion of arctanx fails?

In summary: The series for 1/(1+u^2) has radius of convergence 1 so the series for its square has radius of convergence 1 also. The series for its square should have the same radius of convergence also.In summary, the conversation discusses the use of a power series to find the integral of arctan x, which leads to a question about the validity of the method used. The mistake is identified and the correct series for 1/(1+u^2) is given, leading to the conclusion that the method is not falsifying the result.
  • #1
tamtam402
201
0

Homework Statement


arctan x = ∫du/(1+u2), from 0 to x


Homework Equations





The Attempt at a Solution



I noticed that 1/(1+u2) = 1/(1+u2)1/2 × 1/(1+u2)1/2.

I decided to take the taylor series expansion of 1/(1+u2)1/2, square the result and then integrate.

I got 1/(1+u2)1/2 = 1 - u2 + 3u4/8 - 15u6/48...

When squaring that result, I get 1/1+u2 = 1 - u2 +5u4/8 ...


When I integrate, the first 2 terms are fine but the x5/8 should be x5/5.


Did I make a mistake, or is my method of "cheating" the interval of convergence for the binomial series somehow falsifying the result?
 
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  • #2
tamtam402 said:

Homework Statement


arctan x = ∫du/(1+u2), from 0 to x

Homework Equations



The Attempt at a Solution



I noticed that 1/(1+u2) = 1/(1+u2)1/2 × 1/(1+u2)1/2.

I decided to take the taylor series expansion of 1/(1+u2)1/2, square the result and then integrate.

I got 1/(1+u2)1/2 = 1 - u2 + 3u4/8 - 15u6/48...
I get something different when I square 1 - u2 + 3u4/8 - 5u6/16...

1-2 u2+(7 u4)/4-(11 u6)/8 + ...

How did you get the following?
When squaring that result, I get 1/1+u2 = 1 - u2 +5u4/8 ...


When I integrate, the first 2 terms are fine but the x5/8 should be x5/5.


Did I make a mistake, or is my method of "cheating" the interval of convergence for the binomial series somehow falsifying the result?
 
  • #3
SammyS said:
I get something different when I square 1 - u2 + 3u4/8 - 5u6/16...

1-2 u2+(7 u4)/4-(11 u6)/8 + ...

How did you get the following?

Dangit, I didn't copy my notebook properly sorry.

I squared (1-u2/2 + 3u4/8...), which should be the right series expansion for (1+u2)-1/2.
 
  • #4
For [itex]\left(1 - u^2/2 + 3u^4/8 - 15u^6/48+35u^8/128+...\right)^2[/itex]

I get 1 - u2 + u4 - u6 +u8 + ...
 
  • #5
SammyS said:
For [itex]\left(1 - u^2/2 + 3u^4/8 - 15u^6/48+35u^8/128+...\right)^2[/itex]

I get 1 - u2 + u4 - u6 +u8 + ...

Sigh, I keep making stupid mistakes. I counted u4/2 twice and 3u4/8 once instead of u4/2 + 3u4/8 +3u4/8.

Thank you for the help.
 
  • #6
Rather than using a square root, it would make much more sense to use that the fact that the sum of a geometric series is given by
[tex]\sum_{n=0}^\infty r^n=1+ r+ r^2+ \cdot\cdot\cdot+ r^n+ \cdot\cdot\cdot= \frac{1}{1- r}[/tex]

Here the function is [itex]1/(1+ u^2)[/itex] which is of the form [itex]1/(1- r)[/itex] with [itex]r= -u^2[/itex] so the sum is [itex]1- u^2+ u^4- u^6+\cdot\cdot\cdot= \sum_{n=0}^\infty (-1)^n u^{2n}[/itex]. And the integral of that is [itex]u- (1/3)u^3+ (1/5)u^5- (1/7)u^7+ \cdot\cdot\cdot= \sum_{n=0}^\infty ((-1)^n/(2n+1)!)u^{2n+1}[/itex].
 
  • #7
tamtam402 said:
is my method of "cheating" the interval of convergence for the binomial series somehow falsifying the result?

It is hard to cheat using power series. As long as you do your algebra correctly, the result should be valid in some interval as long as each series in the calculation has a nonzero radius of convergence.
 
Last edited:

FAQ: Where does this method of constructing the taylor expansion of arctanx fails?

Can the Taylor expansion of arctanx be used for all values of x?

No, the Taylor expansion of arctanx fails for x values greater than or equal to 1 or less than or equal to -1. This is because the derivative of arctanx becomes infinite at these points, making the Taylor series diverge.

What is the reason for the failure of the Taylor expansion of arctanx at certain x values?

The Taylor expansion of a function is based on the derivatives of the function at a specific point. In the case of arctanx, the derivative becomes infinite at certain points, causing the Taylor series to diverge and fail to accurately represent the function.

Can the Taylor expansion of arctanx be used to approximate the function at points where it fails?

No, the Taylor expansion of arctanx fails to accurately approximate the function at the points where it fails. This is because the Taylor series is based on the derivatives of the function, and at the points of failure, the derivatives become infinite.

Are there other methods for constructing the Taylor expansion of arctanx that do not have this issue?

Yes, there are other methods such as the Maclaurin series or the binomial series that can be used to construct the Taylor expansion of arctanx without the issue of failure at certain points. These methods involve using different approximations and can sometimes have a wider range of convergence.

Can the Taylor expansion of arctanx be used for other trigonometric functions?

No, the Taylor expansion of arctanx is specific to the arctan function. Other trigonometric functions have their own unique Taylor series that must be used to approximate them. Attempting to use the Taylor expansion of arctanx for other trigonometric functions will result in inaccurate approximations.

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