Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I happen to read a book, in which it states that,

if $P_n(x)$ is the $n+1$ th Taylor polynomial (that is, $P_n(x)=a_0+a_1x+\cdots+a_{n}x^n$) of the function $\sqrt{x+a}$ $(a>0)$at $x=0$, then there is a polynomial $Q_n(x)$ such that


I have tried to calculate for small $n$, it shows the claim is right.

But I cannot find an easy way to prove it, or understand the result clearly. And is there a more general result for other functions, like $\sqrt[3]{x+a}$?

share|cite|improve this question
You know the binomial theorem? – J. M. Aug 29 '11 at 14:46
@J. M.: I think that's unnecessarily specific -- this is true simply because of the degree of the error term. – joriki Aug 29 '11 at 14:53
I was thinking about how s/he'd derive $Q$, but you're spot on, @joriki. (meta: Why'd you delete your answer?) – J. M. Aug 29 '11 at 14:55
Yes, but I can't prove that coefficients of terms $x^k$($1<k\le n$) are zero. – NGY Aug 29 '11 at 14:57
@J. M.: Because in trying to make it look nicer I'd swapped the roles of $\sqrt{x+a}$ and $P_n(x)$ without noticing that squaring would then leave a term with $\sqrt{x+a}$, so I deleted it while fixing that. – joriki Aug 29 '11 at 14:57
up vote 3 down vote accepted

Since $P_n$ is the $(n+1)$-th Taylor polynomial of the function, the error term is of order $x^{n+1}$:


with $R(x)\in O(x^{n+1})$. Squaring that yields


with $S(x)\in O(x^{n+1})$. But $S(x)$ is the difference of two polynomials, and thus itself a polynomial, so if it's in $O(x^{n+1})$ it must be of the form $-x^{n+1}Q_n(x)$ with $Q_n(x)$ a polynomial.

share|cite|improve this answer
Thanks for your answer. I have not tried this way before(I just tried to calculate the coefficients of terms of $P_n(x)^2-x-a$, but it does not work). – NGY Aug 29 '11 at 15:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.