Question

2.

a) Find Ts(x), the third degree Taylor polynomial about x -0, for the function e2

b) Find a bound for the error in the interval [0, 1/2]

3. The following data is If all third order differences (not divided differences) are 2, determine the coefficient of x in P(x). prepared for a polynomial P of unknown degree P(x) 2 1 4

I need help with both. Thank you.

Answer 1

Solution:2 (a) Let
f(x) = er
.

The general form of a Taylor expansion centered at of a function x
f (r)
is

f) (a) n!

. where
(n)(r
is the nth derivative of f(x).

We have to find the third degree Taylor polynomial about $x=0$, for the function
f(x) = er
.

Therefore, substituting a=0 in (i), we have

7l fin)

f(3) (0) 31

31

Since
f(x) = er
,

Therefore

$\Rightarrow f(0)=e^{0}=1,f'(0)=2e^{0}=2, f''(0)=2^2=4,f^{(3)}(0)=2^3=8$

Therefore, (ii) becomes

$e^{2x}=1+2x+\frac{x^{2}}{2!}4+\frac{8}{3!}x^{3}$

$\Rightarrow e^{2x}=1+2x+2x^2+\frac{4}{3}x^{3}$

Therefore, the required third degree Taylor polynomial about $x=0$, for the function
f(x) = er
.

is given by $P_{3}(x)=1+2x+2x^2+\frac{4}{3}x^{3}$

b) Bound for the error in the interval [0, 1/2].

The error term, $R_{n}$ is given by $R_{n}=\frac{f^{(n)}(c)}{(n+1)!}x^{(n+1)}$ for some
c E 0,1/2
.

Now $\frac{d^{n}(e^{2x})}{dx^{n}}=2^{n}e^{2x}$ .

In the given case, n=3 and according to Lagrange's error term, we need n+1=3+1=4 derivative of

f(x) = er
.

That is $\frac{d^{4}(e^{2x})}{dx^{4}}=2^{4}e^{2x}=f^{(4)}(x)$ .

Therefore, the error is given by

$R_{3}=\frac{f^{(4)}(c)}{4!}x^{4}=\frac{x^{4}}{4!}2^4e^{2c}$for some
c E 0,1/2
.

Now we want to find the largest possible value of this error on the interval [0,1/2]. Looking at the equation of $R_{n}$ we see this will happen when the numerator is maximized and the denominator is minimized, i.e. x=1/2 and c=0. Thus an upper bound for $|R_{n}|$ is:

$|R_{n}|=\frac{(1/2)^{4}}{4!}2^4e^{0}=\frac{1}{24}$.

Therefore upper bound for the error in the interval [0, 1/2] is 1/24.