1. ## Evaluate efficiently

This is a question from my numerical analysis class. I apologize in advance if I'm in the wrong thread...

Evaluate
p(x) = 1 - x^3/3! + x^6/6! - x^9/9! + x^12/12! - x^15/15!
as efficiently as possible. How many multiplications are necessary? Assume all coefficients have been computed and stored for later use.

So I know the answer is 7 multiplications by looking in the back of the book. I'm almost positive Horner's Method is needed to show this since it's a big part of this section in my book...

I thought to let y = x^3 and rewrite. I messed around with substitutions of this sort but could never get 7 multiplications. I'm actually not even sure if that's the way to go about it.

If someone could show this, it will be greatly appreciated!! Thanks

2. ## Re: Evaluate efficiently

Originally Posted by jzellt
This is a question from my numerical analysis class. I apologize in advance if I'm in the wrong thread...

Evaluate
p(x) = 1 - x^3/3! + x^6/6! - x^9/9! + x^12/12! - x^15/15!
as efficiently as possible. How many multiplications are necessary? Assume all coefficients have been computed and stored for later use.

So I know the answer is 7 multiplications by looking in the back of the book. I'm almost positive Horner's Method is needed to show this since it's a big part of this section in my book...

I thought to let y = x^3 and rewrite. I messed around with substitutions of this sort but could never get 7 multiplications. I'm actually not even sure if that's the way to go about it.

If someone could show this, it will be greatly appreciated!! Thanks
The Horner method for the computation of a a polynomial of the form...

$p(x)= c_{0}\ x^{n} + c_{1}\ x^{n-1}+...+c_{n}$ (1)

... consists in the computation of the n-th term of the recursive relation...

$p_{k+1}= x\ p_{k}+c_{k+1}\ ;\ p_{0}=c_{0}$ (2)

In this case n=5 and in (2) You must use $x^{3}$ instead of $x$...

Kind regards

$\chi$ $\sigma$