I know the limit rules but that's not helping me out much here this seems so simple but I don't even know where to start. I read the chapter on this and they don't do any examples like this. I believe a sub n converges to a.
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Can you show that $a_n^k\to a^k$ for all $k$? I would suggest to start with that. – Andrés E. Caicedo Mar 11 '14 at 21:25
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do you already know anything about continuity? this would make things much less technical here. – Max Mar 11 '14 at 21:45
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@ Max Yes , I know about continuity. @ Andrew so its like a proof by induction? – Achilles Mar 11 '14 at 21:55
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Clearly, (as you know continuity), the polynomial $P$ is a continuous function mapping from $\mathbb{R}$ to $\mathbb{R}$.
Sequential continuity coincides with continuity in this case, then $a_n\to a$ implies $P(a_n)\to P(a)$.
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OK, check (as you mentioned, by induction, or whatever method you prefer) that for all $k$, $a_n^k\to a^k$. For this, use that the limit of a product of the product of the limits.
Now, if $a_n^k\to a^k$, then also $ca_n^k\to ca^k$ for any constant $c$.
Finally, note that any polynomial is a sum of terms of that form, and use that the limit of a sum is the sum of the limits.
More carefully, if $P(x)=b+cx+dx^2+\dots+ex^m$, then $$P(a_n)=b+ca_n+da_n^2+\dots+ea_n^m,$$ and each term $ra_n^k$ converges to $ra^k$. So $P(a_n)$, being a sum of these terms, converges to $$b+ca+da^2+\dots+ea^m=P(a).$$
Andrés E. Caicedo
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If you know the Ruffini identity, which is connected to the Horner scheme, then you know that there exists a polynomial $q(a,x)$ such that $$ p(x)=p(a)+(x-a)q(a,x) $$ Now you need only to argue that $q(a,x)$ is bounded if the distance from $x$ to $a$ is bounded to get the estimate $$ |p(x)-p(a)|\le M(r)\,|x-a| \quad\text{for}\quad |x-a|<r. $$
Lutz Lehmann
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One elementary way of verifying that there is such a polynomial $q$ is to note that $x^k-a^k=(x-a)(x^{k-1}+x^{k-2}a+\dots+xa^{k-2}+a^{k-1})$, so it is $(x-a)$ times a polynomial in $x$ and $a$. This is for each $k$, and $p(x)-p(a)$ is a sum of multiples of these differences. I thought about including this argument, but I suspect that the task of showing boundedness may be seen about as hard as the original question. – Andrés E. Caicedo Mar 11 '14 at 22:26
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No, the bound is easy, just apply the triangle inequality to $|q(a,x)|$ to replace everything by their absolute values and then replace $|x|$ by $|a|+r$ to get the constant $M(r)$. -- If Horner-Ruffini is not known, then your answer shows the easier way. – Lutz Lehmann Mar 11 '14 at 22:54
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I know it is easy, but then again, so is the question at hand. They feel comparable, is what I meant. By the way, I didn't know the identity was named after Ruffini. – Andrés E. Caicedo Mar 11 '14 at 23:09
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There is a rather silly discussion on the wikipedia talk pages, if the two articles Ruffini's rule and Horner scheme should be merged or if they really are different topics. What I understand as a reasonable point of view is that the synthetic division formula is named for Ruffini, and the computation scheme for Horner, but there is a large area of overlapping content. The Horner article is by far the better written one. – Lutz Lehmann Mar 11 '14 at 23:34