Question

Starting from the geometric series, use power series operations to

determine the Maclaurin series. See picture

Answer 1

a. By replacing
`x` with
`-x^2` in the power series, we get

`\displaystyle \frac1{1+x^2} = \sum_(n=0)^\infty (-x^2)^n = \sum_(n=0)^\infty (-1)^n x^(2n)`

Integrate both sides to recover
`\arctan(x)` on the left.

`\displaystyle \int (dx)/(1+x^2) = \int \sum_(n=0)^\infty (-x^2)^n = \sum_(n=0)^\infty (-1)^n x^(2n) \, dx`

`\displaystyle \arctan(x) = C + \sum_(n=0)^\infty ((-1)^n)/(2n+1) x^(2n+1) \, dx`

By letting
`x=0` on both sides, we find
`C=0`, so that

`\displaystyle \arctan(x) = \sum_(n=0)^\infty ((-1)^n)/(2n+1) x^(2n+1) \, dx`

Then dividing both sides by
`x` gives

`\displaystyle \boxed{\frac{\arctan(x)}x = \sum_(n=0)^\infty ((-1)^n)/(2n+1) x^(2n) \, dx}`

b. Let
`x=\frac1{\sqrt3}`. Then

`\displaystyle \frac{\arctan\left(\frac1{\sqrt3}\right)}{\frac1{\sqrt3}} = \sum_(n=0)^\infty ((-1)^n)/(2n+1) \left(\frac1\sqrt3\right)^(2n)`

`\displaystyle \sqrt3 \arctan\left(\frac1{\sqrt3}\right) = \sum_(n=0)^\infty \frac1{2n+1} \left(-\frac13\right)^n`

Taking the first few terms from the infinite series, we can approximate

`n=0 \implies \sqrt3\arctan\left(\frac1{\sqrt3}\right) \approx 1`

`0\le n\le1 \implies \sqrt3\arctan\left(\frac1{\sqrt3}\right) \approx 1+\frac13\left(-\frac13\right) = \frac89`

which together suggest the value we want is bounded between 8/9 and 9/9 = 1, hence
`\boxed{p=8}`.

Since the series is alternating and converges on
`-1<x<0\cup0<x<1`,

`\displaystyle \left| \sum_(n=0)^\infty a_n - \sum_(n=0)^0 a_n \right| < |a_1| = \left|\frac13 \left(-\frac13\right)^1\right| = \frac19`

and

`\displaystyle \left| \sum_(n=0)^\infty a_n - \sum_(n=0)^1 a_n \right| < |a_2| = \left|\frac15 \left(-\frac13\right)^2\right| = \frac1{45}`

which tells us the first approximation is off by at most 1/9 from the actual value of
`\frac\pi{2\sqrt3}`, whereas the second approximation is off by at most 1/45 from the actual value. In other words, the second approximation is closer, so
`\frac\pi{2\sqrt3}` is closer to
`\frac p9` :

`\frac89 \approx 0.8889`

`(\pi)/(2\sqrt3)\approx0.9069`

`\frac99 = 1`