Question

How do you find the first two nonzero terms in Maclaurin's Formula and use it to approximate `f(1/3)` given `f(x)=arcsinx`?

Answer 1

Knowing that `d/dxarcsinx=1/sqrt(1-x^2)`, let's try to find the Maclaurin series first for `(1-x^2)^(-1/2)`.

Note the binomial series:

`(1+z)^alpha=sum_(k=0)^oo((alpha),(k))z^k`

Where `((alpha),(k))-=(alpha(alpha-1)(alpha-2)...(alpha-k+1))/(k!)`.

So, when `z=-x^2` and `alpha=-1/2`, we see that

`k=0=>((-1/2),(0))(-x^2)^0=1/(0!)=1`

`k=1=>((-1/2),(1))(-x^2)^1=(-1/2)1/(1!)(-x^2)=x^2/2`

`k=2=>((-1/2),(2))(-x^2)^2=(-1/2)(-3/2)1/(2!)x^4=(3x^4)/8`

Continuing this process, if you wish, gives:

`1/sqrt(1-x^2)approx1+x^2/2+(3x^4)/8+(5x^6)/16+(35x^8)/128+...`

Then,

`arcsinx=intdx/sqrt(1-x^2)=int(1+x^2/2+(3x^4)/8+(5x^6)/16+(35x^8)/128+...)dx`

Using the first two nonzero terms, this gives:

`arcsinxapproxx+x^3/6`

Then,

`arcsin(1/3)approx1/3+(1//3)^3/6=1/3+1/(162)=0.339506...`

Compare this to the more exact value, `arcsin(1/3)=0.339837...`