Recent content by jimpap

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.

Yes it was quite explanatory , I see that the degree is <=k.

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!

You mean to try different k for: f(x)=ak+1xk+1+akxk+...+a1x+a0, and h(x)=f(x)-ak+1xk+1 or h(x)=f(x)-ak+1(x-a)xk

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...

And again thank you for the suggestions. Both seem very interesting!

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.

I am 30 years old and I thought to refresh my math knowledge. I am glad to be part of this forum, seems very active! Regards