2
$\begingroup$

I have searched and read quite a bit on this subject but I can't get this last bit straight. Reading the other answers did not help me unfortunately for me.

Anyway the problem: Suppose I have the function $g(x) = (8+3x)^{\frac{1}{3}}$ and I want to use Taylor expansion around $x = 0$ to find the third degree polynomial $P_3(x) = 2 + \frac{x}{4} - \frac{x^{2}}{32} + \frac{5x^{3}}{768}$. I shall then use this to find an approximation for $11^{\frac{1}{3}}$..

I found $g(x) = 11^{\frac{1}{3}}$ when $x = 1$. Then I'm supposed to find the remainder/error $E_3(x)$ and use this to find a small interval that will certainly contain the correct value $11^{\frac{1}{3}}$.

I want to use this formula $E_n(x) = \dfrac{g^{(n+1)}(c)(x-a)^{n+1}}{(n+1)!}$

to find $E_3(x)$ .. So I went ahead and found $g^{(4)}(x) = -\dfrac{80}{(8+3x)^{\frac{11}{3}}}$

Which means $E_3(x) = g^{(4)}(c)\dfrac{x^{4}}{4!}$

Now my problem is choosing $c$ and that's what I wanted to ask about. I don't know exactly what I'm doing here. Is it true that I must choose $c \in[0, 1]$ because I centered the series around $x = 0$, and $1$ because $g(1) = 11^{\frac{1}{3}}$ ? Also when choosing $c$, is it true that I want the one that returns the biggest value so the error won't be too small?

In that case I guess I must pick $c = 1$ ?

I hope someone can give an answer that is easy to understand, because im really not confident in this subject and dont't have the deep understanding(yet :)), also I'm quite new to calculus.

$\endgroup$
2
  • 1
    $\begingroup$ note that $g^{(n)}$ and $g^n$ is not the same thing, one is a derivative and another is a power $\endgroup$ Commented Oct 23, 2015 at 0:46
  • $\begingroup$ For pessimistic bound, which is the best you can do, assume $c=0$. $\endgroup$ Commented Oct 23, 2015 at 0:59

2 Answers 2

1
$\begingroup$

We know nothing about $c$, except that it is between $0$ and $1$, since we are expanding about $0$ and evaluating at $1$.

Note that the absolute value of $-\frac{80}{(8+3x)^{11/3}}$ decreases as $x$ increases from $0$ to $1$. Since conceivably $c$ might be very near $0$, and we want to give a guarantee that the error is no greater than our bound, we are forced to choose the most pessimistic estimate for $c$, which is $c=0$.

$\endgroup$
3
  • $\begingroup$ You are welcome. The Lagrange error term is in principle exact, apart from the fact "$c$" is not known. Thus we often don't get good error estimates from it. Adding to the problem is that higher derivatives can be quite complicated. As a result, it is seldom a first choice for error estimates in real situations. $\endgroup$ Commented Oct 23, 2015 at 5:11
  • $\begingroup$ You cannot "choose" c.Its value,( or allowable values , when the (n+1)th derivative is not 1-to-1) is dictated by the equation :"There exists c such that...." $\endgroup$ Commented Oct 23, 2015 at 5:18
  • $\begingroup$ @user254665: I altered the wording somewhat. $\endgroup$ Commented Oct 23, 2015 at 5:34
1
$\begingroup$

You shouldn't guess $c$, but you can estimate the error, i.e. you take the $c\in[x_0,x]$ ($f(x)=f(x_0)+...$) that give a worse case for $g^{(n)}$, i.e. biggest value, and show that your error is less then something.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .