Approximating square root of 2 (Taylor remainder)

[Poetria]

Homework Statement

[/B]
Use the Taylor remainder theorem to give an expression of
$\sqrt 2 - P_3(1)$

P_3(x) - the degree 3 Taylor polynomial $\sqrt {1+x}$ in terms of c, where c is some number between 0 and 1

Find the maximum over the interval [0, 1] of the absolute value of the expression you entered above.2. The attempt at a solution

The fourth derivative of $\sqrt {x}$ :
$-\frac {15} {16*(x)^\frac 7 2}$

$\frac{-\frac {15} {16*(c)^\frac 7 2}} {4!}*x^4$
x=1
$\frac{-\frac {15} {16*(c)^\frac 7 2}} {4!}*1^4$

for c = 1 and x = 1

$|\sqrt 2 - P_3(1)| \leq |-5/128|$

 

[Mark44]

Homework Statement

[/B]
Use the Taylor remainder theorem to give an expression of
$\sqrt 2 - P_3(1)$

P_3(x) - the degree 3 Taylor polynomial $\sqrt {1+x}$ in terms of c, where c is some number between 0 and 1

Find the maximum over the interval [0, 1] of the absolute value of the expression you entered above.2. The attempt at a solution

The fourth derivative of $\sqrt {x}$ :
$-\frac {15} {16*(x)^\frac 7 2}$

$\frac{-\frac {15} {16*(c)^\frac 7 2}} {4!}*x^4$
x=1
$\frac{-\frac {15} {16*(c)^\frac 7 2}} {4!}*1^4$

for c = 1 and x = 1

$|\sqrt 2 - P_3(1)| \leq |-5/128|$

I agree with your answer of 5/128, but not with some of your work. Following your work wasn't as easy as it could be, since it wasn't clear to me whether the series you were working with was a Taylor series (i.e., in powers of x - a) or a Maclaurin series, in powers of x. From your work you are apparently writing the Maclaurin series for $f(x) = \sqrt{1 + x}$.

In calculating the remainder term, you aren't working with the same function. IOW, for the remainder term, the function should still be $f(x) = \sqrt{1 + x}$, not $f(x) = \sqrt x$. The remainder term will be $\frac{f^{4}(c)}{4!}x^4$, evaluated at x = 1, for some c between 0 and 1. You need to convince yourself (and your teacher) that the fourth derivative term is largest when c = 0, and not at some other value in the interval (0, 1).

 

[Poetria]

I agree with your answer of 5/128, but not with some of your work. Following your work wasn't as easy as it could be, since it wasn't clear to me whether the series you were working with was a Taylor series (i.e., in powers of x - a) or a Maclaurin series, in powers of x. From your work you are apparently writing the Maclaurin series for $f(x) = \sqrt{1 + x}$.

In calculating the remainder term, you aren't working with the same function. IOW, for the remainder term, the function should still be $f(x) = \sqrt{1 + x}$, not $f(x) = \sqrt x$. The remainder term will be $\frac{f^{4}(c)}{4!}x^4$, evaluated at x = 1, for some c between 0 and 1. You need to convince yourself (and your teacher) that the fourth derivative term is largest when c = 0, and not at some other value in the interval (0, 1).

I wasn't sure if there should be sqrt(x) or sqrt(1+x). Thank you very much. Now I understand everything. :) :) :)