analysiscalculusreal-analysissequences-and-seriestaylor expansion
Something is bothering me with the remainder of the Taylor (Maclaurin) series of $\cos(x)$.
The formula of $a_n$ is $(-1)^n \frac{x^{2n}}{(2n)!}$.
By Leibniz Theorem, $r_n<a_{n+1}$ which is, $\frac{x^{2n+2}}{(2n+2)!}$. The Lagrange remainder is $r_n= \frac{f^{(2n+1)}(c) \cdot x^{2n+1}}{(2n+1)!}$. My question is, knowing that $f^{2n+1}(x)=0$ when $x=0$. Can we say that $r_n=r_{n+1}$ and take the lagrange remainder to be $r_n= \frac{f^{(2n+2)}(c) \cdot x^{2n+2}}{(2n+2)!}$?
Thank you,
Shir
I see that this is a pretty old question, but here goes anyway.
The main problem with a geometric approach is that we are often dealing with very high order derivatives. Just by looking at a graph we can easily get a sense of such geometric interpretations as value, gradient and concavity - corresponding respectively to $f^{(0)}(x)$, $f^{(1)}(x)$ and $f^{(2)}(x)$ - but after that it starts to become a struggle to interpret function behaviour visually.
Having said that, if we choose $f(x)=e^x$, for which $f^{(k)}(x) = e^x$ for all $k$, we can produce something of a geometric interpretation of the Lagrange error term, especially if we start off with low degree Taylor polynomials for the approximation and incrementally incorporate more terms from the series into the polynomial approximation.
We know that:
$$\begin{align} f(b) = \sum_{k=0}^\infty\frac{f^{(k)}(a)(b - a)^k}{k!}&= \sum_{k=0}^n\frac{f^{(k)}(a)(b - a)^k}{k!}+\sum_{k=n+1}^\infty\frac{f^{(k)}(a)(b - a)^k}{k!}\\\\ &= T_n(b:a) + R_n(b:a) \end{align}$$
And that:
$$\left(\exists c \in ]a, b[\right)\left(\frac{f^{(n+1)}(c)(b-a)^{n+1}}{(n+1)!}=R_n(b:a)=\sum_{k=n+1}^\infty\frac{f^{(k)}(a)(b - a)^k}{k!}\right)$$
Now, for example, let's say that we want to use a Taylor series of $f(x) = e^x$ about $a = 0$ to estimate $f(b = 2)$.
If we make the $n=0$ assertion (in other words, we say that $f(b)\approx f^{(0)}(a)$ independent of $b$, which generally is only a good idea for $b$ very close to $a$), we are effectively saying that there is some $c$ in the region $]a, b[$ such that:
$$\frac{f^{(0+1)}(c)(b-a)^{0+1}}{(0+1)!}=e^b - e^a$$
In this particular example, where we have specified the values, we can calculate $c$:
$$2e^c=e^2 - 1\\c = \ln\left(\frac{e^2 - 1}{2}\right)\approx 1.16$$
With $c$ and $f(x)$ translated down by twice the gradient of $f$ at $c$ (which, since we're using $f(x)=e^x$, is twice $f(c)$) shown:
Similarly, if we make the $n=1$ assertion (in other words we say that $f(b)\approx f^{(0)}(a)+ f^{(1)}(a)(b - a)$) we are effectively saying that there is some $c$ in the region $]a, b[$ such that:
$$\frac{f^{(1+1)}(c)(b-a)^{(1 + 1)}}{(1+1)!}=e^b - (e^a + e^a(b - a))$$
Since we have specified all values we can again calculate $c$, which, together with the graph of $f(x)$ translated down by twice the concavity of $f$ at $c$ (which, since we're using $f(x) = e^x$, is twice the value of $f$ at $c$) is illustrated:
Of course, we can continue on similarly from here (though we have to now pay a little more attention to the factorials in the denominators), incorporating successively more terms from the Taylor series into the Taylor polynomial. The nice thing about using $f(x)=e^x$ is that we will always be able to interpret derivatives evaluated at $c$ as the height of the curve above the $x$ axis at $c$, which would not be the case with other functions.
While a geometric interpretation of the Lagrange error term would be a lot more complicated with other functions, I found that when I could make sense of what was happening in the simple $e^x$ case, the Lagrange error term made a lot more sense in general.
If the remainder term $R_N$ does not converge to $0$ then the power series will not sum to $f(x).$
For just the case $0\leq x<1/2 :\quad$ Let $x=1/2-y.$ Then $c_N$ (which you write as $c$) satisfies $1-c_N>1-x$ for every $N. $ Let $z=x/(1-x).$ Then $0\leq z<1. $
For every $N$ we have $$0<x/(1-c_N)<z<1.$$ Note that $z$ does not depend on $N.$
Now since $0\leq z<1, $ there exists $k$ with $0<k<1$ such that for all sufficiently large $N$ we have $z(1+3/(2 N+2))<k.$ So for all sufficiently large $N$ we have $0<R_{N+1}<k R_N,$ so $R_N\to 0.$
That is, choose $N_0$ such that $n> N_0\implies z(1+3/(2 n+2))<k. $ Then for all $m> 0$ we have $0<R_{N_0+m}<k^m R_{N_0}.$
Best Answer
The textbook "An introduction to numerical methods and analysis" by James F. Epperson, 2nd edition, page 3:
$$ \cos x = 1 - \frac{1}{2!}x^2 + \frac{1}{4!}x^4 + \dotsm + \frac{(-1)^n}{(2n)!}x^{2n} + \frac{(-1)^{n+1}}{(2n+2)!}x^{2n+2} \cos \xi_x $$
So the answer is yes, because the first term for remainder is zero then you can go for the next n.