Taylor series of arctan(x) (Spivak)

At p. 388 of Calculus, Spivak gives a formula:

$$\frac{1}{1+t^2} = 1 - t^2 + t^4 - ... + (-1)^nt^{2n} + \frac{(-1)^{n+1}t^{2n+2}}{1+t^2}$$

Which can be integrated to find $\arctan(x)$.

I don't understand where this formula comes from, but I found it up to $(-1)^nt^{2n}$ by considering the geometric series for $\frac{1}{1-x}$ and replacing $x$ by $-x^2$ to get the series for $\frac{1}{1+x^2}$. I don't see the term $\frac{(-1)^{n+1}x^{2n+2}}{1+x^2}$ though, because the series I got this way is $\frac{1}{1+x^2} = \sum_{n=0}^{\infty}(-1)^nx^{2n}$.

• In fact, Taylor series has a remainder, this remainder can be written in an integration form or big O form, and as in this case it is written in a fraction form . – Nizar Oct 5 '15 at 9:14

Consider $$1+x+x^2+\dots+x^n=\frac{1-x^{n+1}}{1-x}=\frac{1}{1-x}-\frac{x^{n+1}}{1-x}$$ that can also be written $$\frac{1}{1-x}=1+x+x^2+\dots+x^n+\frac{x^{n+1}}{1-x}$$ Now substitute $x=-t^2$, that gives $$\frac{1}{1+t^2}=1+(-t^2)+(-t^2)^2+\dots+(-t^2)^n+\frac{(-t^2)^{n+1}}{1+t^2}$$ and not it's just a matter of observing that $(-t^2)^k=(-1)^kt^{2k}$.

• This is a simple derivation since it builds from elementary series. – mavavilj Oct 5 '15 at 20:38

The term $(-x^2)^{n+1}/(1+x^2)$ is just the rest term:

$${1\over 1 - (-x^2)} = \sum_0^\infty (-x^2)^k = \sum_0^n (-x^2)^k + \sum_{n+1}^\infty (-x^2)^k$$

where

$$\sum_{n+1}^\infty (-x^2)^k = (-x^2)^{n+1} \sum_0^\infty (-x^2)^k = {(-x^2)^k\over1 - (-x^2)}$$

Alternately you can do it the other way from the RHS and use the formula for geometric series (so you won't have to resort to infinite series):

$$\sum_0^n(-x^2)^k + {(-x^2)^{n+1}\over 1 - (-x^2)} = {1 - (-x^2)^{n+1}\over 1- (-x^2)} + {(-x^2)^{n+1}\over 1 - (-x^2)} = {1\over 1+x^2}$$

• Nicely done. The equations would be easier to read in display mode; use \$\$ instead of \$at the start and end of each large equation. – David K Oct 5 '15 at 13:57 • I believe technically one cannot claim$\sum_0^\infty (-x^2)^k = \sum_0^n (-x^2)^k + \sum_{n+1}^\infty (-x^2)^k$without knowing first that$\sum_0^\infty (-x^2)^k$has a sum / converges. Since if it doesn't (e.g. diverges to$\infty$) then$\sum_0^n (-x^2)^k + \sum_{n+1}^\infty (-x^2)^k$does not convey anything, since the sums may be arbitrary (e.g.$\infty$). – mavavilj Oct 5 '15 at 20:36 • @mavavilj That's correct, you have to make sure that the sum actually converges, which is true if$|x| < 1$. – skyking Oct 5 '15 at 21:13 This is a finite geometric sum: $$\sum_{k=0}^n (-1)^k t^{2k} = \sum_{k=0}^n (-t^2)^k = \frac{1-(-1)^{n+1}t^{2n+2}}{1-(-t^2)}$$ • I know the geometric sum formula, but what about the other terms preceding$\frac{(-1)^{n+1}t^{2n+2}}{1+t^2}$in Spivak's formula? – mavavilj Oct 5 '15 at 9:09 • @mavavilj What do you mean? The other terms are$\sum_{k=0}^n (-t^2)^k\$. Just rearrange the above equality to get exactly what you have written. – mrf Oct 5 '15 at 9:31