Appreciate your interest. I think spivak's book is pretty difficult. What I was asking clarified at https://www.physicsforums.com/threads/polynomial-division-continued-from-osnarfs-problem.842252/#post-5285189.
And again thank you.
I see that the degree is the same as number k and ak+1xk+1 is simplified.
But since ak+1xk+1 is irrelevant to ak+1(x-a)xk, against my original thought,
I can't understand how should I think the h(x) polynomial.
Thank you for your time!
If it not claimed that this is the same as f(x) minus its leading term, did we take
the polynomial h(x)=f(x)-a[k+1](x-a)x[k], meaning
that f(x) a random polynomial of k+1 power and subtract another polynomial also to the k+1 power?
So the a[k+1]x[k+1] has nothing to do with a[k+1](x-a)x[k]?
Sorry, I still miss something.
Yes it became degree of k at most because we subtracted a[k+1]x[k+1].
So, we should have :
f(x)-a[k+1]x[k+1] = (x-a)g(x) + b, is this correct, since we have degree at k the most?
And then
a[k+1]x[k+1] = a[k+1](x-a)x[k] and we don't take into consideration
the...
Thanks for your instant response.
But :
a[k+1](x-a)x[k]
isn't it equal to:
a[k+1]x[k+1] - a[k+1]ax[k].
So, turning from
a[k+1]x[k+1]
to
a[k+1](x-a)x[k]
wouldn't require to add
a[k+1]ax[k]?
Hello,
My problem is the same as osnarf's problem in thread "Polynomial division proof",
https://www.physicsforums.com/threads/polynomial-division-proof.451991/
But, I would like some further help.
The problem:
Prove that for any polynomial function f, and any number a, there is a polynomial...
Thank you,
I 've just started studying "Michael Spivak Calculus" and I find it very difficult. So, pretty much I 'd like some further explanation - help to some of the exercises.