How do you find MacLaurin's Formula for f(x)=sin(2x) and use it to approximate f(12) within 0.01?

1 Answer
Mar 8, 2017

f(12)=42415040+R with |R|<0.0084

Explanation:

Start from the MacLaurin expansion of sint:

sint=n=0(1)nt2n+1(2n+1)!

substitute t=2x

f(x)=sin(2x)=n=0(1)n(2x)2n+1(2n+1)!

So for x=12

f(12)=n=0(1)n(212)2n+1(2n+1)!=n=0(1)n(2n+1)!

Using Lagrange's formula for the error we have:

Rn(12)=[dn+1(dx)n+1sin(2x)]x=c(n+1)!(12)n+1

where 0<c<12.

Now we have:

dn+1(dx)n+1sin(2x)=2n+1sin(x+(n+1)π2)

so as |sinx|1:

∣ ∣[dn+1(dx)n+1sin(2x)]x=c∣ ∣2n+1

and then:

Rn(12)2n+1(n+1)!(12)n+1

that is:

Rn(12)1(n+1)!

Thus we have to determine n such that:

1(n+1)!0.01

which is satisfied for n4

In conclusion we have:

f(12)=116+112015040+R4=5040840+4215040+R4=42415040+R40.84

Where |R4|<1120