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

[tamtam402]

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?

 

[SammyS]

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?

 

[tamtam402]

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.

 

[SammyS]

For $\left(1 - u^2/2 + 3u^4/8 - 15u^6/48+35u^8/128+...\right)^2$

I get 1 - u² + u⁴ - u⁶ +u⁸ + ...

 

[tamtam402]

For $\left(1 - u^2/2 + 3u^4/8 - 15u^6/48+35u^8/128+...\right)^2$

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.

 

[HallsofIvy]

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
$$\sum_{n=0}^\infty r^n=1+ r+ r^2+ \cdot\cdot\cdot+ r^n+ \cdot\cdot\cdot= \frac{1}{1- r}$$

Here the function is $1/(1+ u^2)$ which is of the form $1/(1- r)$ with $r= -u^2$ so the sum is $1- u^2+ u^4- u^6+\cdot\cdot\cdot= \sum_{n=0}^\infty (-1)^n u^{2n}$. And the integral of that is $u- (1/3)u^3+ (1/5)u^5- (1/7)u^7+ \cdot\cdot\cdot= \sum_{n=0}^\infty ((-1)^n/(2n+1)!)u^{2n+1}$.

 

[Vargo]

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.