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 $\frac{1}{1+x^2} = \sum_{n=0}^{\infty}(-1)^nx^{2n}$.

  • $\begingroup$ 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 . $\endgroup$ – 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 $(-t^2)^k=(-1)^kt^{2k}$.

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

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

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


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

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):

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

  • $\begingroup$ Nicely done. The equations would be easier to read in display mode; use \$\$ instead of \$ at the start and end of each large equation. $\endgroup$ – David K Oct 5 '15 at 13:57
  • $\begingroup$ 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. $\infty$). $\endgroup$ – mavavilj Oct 5 '15 at 20:36
  • $\begingroup$ @mavavilj That's correct, you have to make sure that the sum actually converges, which is true if $|x| < 1$. $\endgroup$ – skyking Oct 5 '15 at 21:13

This is a finite geometric sum:

$$ \sum_{k=0}^n (-1)^k t^{2k} = \sum_{k=0}^n (-t^2)^k = \frac{1-(-1)^{n+1}t^{2n+2}}{1-(-t^2)} $$

  • $\begingroup$ 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? $\endgroup$ – mavavilj Oct 5 '15 at 9:09
  • $\begingroup$ @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. $\endgroup$ – mrf Oct 5 '15 at 9:31

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