Math Help!!



#2

Hi All,

How can one convert a polynomial of the general (and common) form:

p(x) = Sum a(i) x^i, for i = 0,1,2,..,n
to a nested form:
p(x) = A(0) + x^B(0)(1 + A(1)x^B(1)(1 + A(2)x^B(2)(1 + A(3)x^B(3)( .....(1 + A(n)x^B(n)))...)

Notice that each open parenthesis is in the form of:

1 + A(i)x^B(i)

I saw these nested polynomials, as approximations for various common log and trig functions, on page 37 in Jon Smith's book Scientific Analysis on the Pocket Calculator, 2nd edition.

Namir

Edited: 16 Apr 2012, 6:56 p.m.


#3

Are you describing Horner's method here ?

If so, start factoring powers of x from whichever end has the highest power.


- Pauli


#4

Its' a variant of the regular Horner's method.

#5

Namir,

I'm not sure why you're looking into this, but this technique was very useful for writing a compact program to calculate pi like I did in this set of programs.

-Katie


#6

Yes Katie, the form of the polynomial in your link is what I am looking for.

Namir

#7

I have that very book. On the page in question Smith has simply applied Horner's method, only in all of the examples on that page the central polynomial of the approximations in question are in x^2 rather than x. If you take a close look, you will see that all of your B(i)'s equal 2.

Hope this helps.

Les


#8

Yes, all B(i) would typically be the same value.

#9

Polynomial Expressions and Horner's Method are discussed on p. 79 of the HP 15C Owner's Handbook, but not How to Do It. Factorization is clearly what's required.

#10

Namir, here is a simple 10th-degree example without odd-powered terms to demonstrate the method Smith uses:

1 + 3x^2 +4x^4 + 5x^6 + 2x^8 + 7x^10 = 1 + x^2(3 + x^2(4 + x^2(5 + x^2(2 + 7x^2))))

If C is a 0-based 5 element vector containing the coefficients in the order written, a C-code snippet might look like this:

C=[1,3,4,5,2,7];
y=x*x;
p=C[n-1];
for(j=n-2;j>=0;j--) p=p*y+c[j];

FWIW, it seems that Smith never uses the term "Horner's Method", but he does discuss the process of rewriting polynomials in this nested form on pages 31-32ff.

HTH,

Les

Edited: 16 Apr 2012, 9:01 p.m.


#11

Yes, just to be clear, the coefficients in both forms are the same. This is not "factoring".

#12

I use wolfram alpha to calculate Horner forms. example:

http://www.wolframalpha.com/input/?i=4*Sum[4*n^2%2Bn%2B1%2C+{n%2C+0%2C%28s-1%29%2F2}]-3

Look under alternate forms.


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