Question

Find the third Taylor polynomial P3(x) for the function f(x) = (x-1)ln(x) about x0 = 1.

(a) Use P3(0.5) to approximate f(0.5). Find an upper bound for error |f(0.5) - P3(0.5)| using the error formula, and compare it to the actual error.
(b) Find a bound for the error |f(x) - P3(x)| in using P3(x) to approximate f(x) on the interval [0.5,1.5].
(c) Approximate \int_{0.5}^{1.5}f(x)dx using \int_{0.5}^{1.5}P_{3}(x)dx .
(d) Find an upper bound for the error in (c) using \int_{0.5}^{1.5}|R_{3}(x)dx| , and compare the bound to the actual error.

Answer 1

Answer:

A) The actual error |f(0.5) - P3(0.5)| is 0.0340735. The error using Taylor's error formula R₃(0.5) is 0.2917.

B) Te error in the interval [0.5,1.5] is 0.0340735.

C) The approximate integral using P₃(x) is 1/12

D) The real error between the two integrals is 4.687·10^(-3) while the error using \int_{0.5}^{1.5}|R_{3}(x)dx| is 0.018595

Step-by-step explanation:

A) To determine the error we first have to write the third grade Taylor polynomial P3(x):

$P_n(x)=\displaystyle\sum_{k=0}^N \frac{f^{k}(a)}{k!}(x-a)$ with N=3

$P_3(x)=\displaystyle\sum_{k=0}^3 \frac{f^{k}(1)}{k!}(x-1)=(x-1)^2-\frac{1}{2} (x-1)^3$

The error for the third grade Taylor polynomial R₃(x) is represented as the next term non-written in the polynomial expression near to the point x:

$R_3(x)=\displaystyle \frac{f^{4}(x)}{4!}(x-a)=\frac{1}{24} (\frac{2}{x^3}+\frac{6}{x^4})(x-1)^4$

Therefore the real error is:

$Err(0.5)=|f(0.5)-P_3(0.5)|=0.0340735$

The error of the Taylor polynomial R₃(0.5) is:

$R_3(0.5)=\displaystyle \frac{1}{24} (\frac{2}{0.5^3}+\frac{6}{0.5^4})(0.5-1)^4=0.2917$

B) The error in the interval [0.5,1.5] is the maximum error in that interval.

This is found in the extremes of the intervals. We analyze what happens in X=1.5:

$Err(1.5)=|f(1.5)-P_3(1.5)|=0.01523$

$R_3(1.5)=\displaystyle \frac{1}{24} (\frac{2}{1.5^3}+\frac{6}{1.5^4})(1.5-1)^4=\frac{1}{216}=4.63\cdot 10^{-3}$

Both of these errors are smaller than Err(0.5) and R₃(0.5). Therefore the error in this interval is Err[0.5,1.5] is Err(0.5).

C) The approximation of the integral and the real integral is:

$I_f=\displaystyle\int_{0.5}^{1.5} f(x)\, dx=\int_{0.5}^{1.5} (x-1)ln(x)\, dx=0.08802039$

$I_{p3}=\displaystyle\int_{0.5}^{1.5} P_3(x)\, dx=\int_{0.5}^{1.5} (x-1)^2-0.5(x-1)^3\, dx=1/12=0.0833333$

D) The error in the integrals is:

$Errint=|I_f-I_{p3}|=4.687\cdot10^{-3}$

$Errint(R_3)=\displaystyle\int_{0.5}^{1.5} R_3(x)\, dx=\int_{0.5}^{1.5} \frac{1}{24} (\frac{2}{x^3}+\frac{6}{x^4})(x-1)^4\, dx=0.018595$