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I'm reading Mathematics: It's Content, Methods, and Meanings and I am in a chapter about Taylor Series. It made sense until I came across the remainder part of the theorem. In order to prove the theorem it uses ${f(x)-P_n(x) \over -(x-a)^{n+1}}$ to be able to use Cauchy's Mean Value theorem to prove the Langrange form of the remainder. Could someone explain why they choose $-(x-a)^{n+1}$. Is it to make the proof easier or is there an underlying meaning? If so please explain why. Also, if it is just to make the proof easier is there a better (more intuitive) alternative proof?

Thanks,

Jackson

EDIT: Here is part of a proof from http://davidlowryduda.com/?p=1804 Proof

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Two things to say here: firstly, $G$ as defined in the proof you give has $G'(c) = -(n+1)(x-c)^{n}$, so the minus signs cancel out, and you have $$ R_n = (x-a)^{n+1} \frac{F'(c)}{(n+1)(x-c)^n}, $$ and presumably the rest of the proof is observing that $F'(c) = (x-c)^n f^{(n+1)}(c)/n!$, giving $$ R_n = \frac{(x-a)^{n+1}}{(n+1)!} f^{(n+1)}(c) .$$ Basically you have to choose it this way if you work with the polynomial $F$ with $(x-t)$ in it.


This is essentially the only way to prove Taylor's theorem with the mean-value remainders (I explained in my answer to a previous question of yours how to derive the integral form). Suppose we use the same $F$. Then the expression we have for the remainder is $$ R_n = \frac{(x-c)^n f^{(n+1)}(c)}{n!} \frac{G(x)-G(a)}{G'(c)}. $$ This is quite general, and you can get a variety of remainders from using different functions. The most basic thing you might try is $G(t) = At+B$. Then $$ G'(c) = A, \quad G(x)-G(a) = A(x-a), $$ so whatever linear function you use, you find $$ R_n = \frac{(x-c)^n f^{(n+1)}(c)}{n!} (x-a) $$ (this is called the Cauchy form of the remainder, in fact). Suppose now we try a quadratic polynomial. The argument above shows that we can just use a monic polynomial, and the constant is irrelevant, and we know about completing the square, so we can try $G(x) = (A-t)^2$ as the most general. $$ G'(c) = -2(A-t), \quad G(x)-G(a) = (x-a)(x+a-2A), $$ and we get $$ R_n = \frac{(x-c)^n f^{(n+1)}(c)}{n!} \frac{(x-a)(2A-x-a)}{2(A-c)}. $$ Well, this is not a very nice expression: but if we set $A=x$, the numerator simplifies to $(x-a)^2$ and the denominator cancels an $x-c$ from the $F'(c)$. You can do similar things with other degrees of polynomial: in general, you can consider $$ G(t) = \sum_{k=0}^{N} A_k (x-t)^k, $$ but the simplest one to take is $(x-t)^k$, which gives you $$ R_n = \frac{(x-c)^{n-k+1} f^{(n+1)}(c)}{n!k} (x-a)^k; $$ finally, taking $k=n+1$ gives the Lagrange form of the remainder, $$ R_n = \frac{f^{(n+1)}(c)}{(n+1)!} (x-a)^{n+1} $$

(Also, from the above discussion we conclude that $G(t) = -(x-t)^n$ would also have been a perfectly legitimate choice for the Lagrange form of the remainder.)

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  • $\begingroup$ Nicely done. +1 $\endgroup$ – Simon S May 22 '15 at 14:58

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