Find the polynomial of lowest degree that will approximate $F ( x )$ throughout the given interval with an error of magnitude less than than $10 ^ { - 3 }$.
-$$\mathrm { F } ( \mathrm { x } ) = \int _ { 0 } ^ { \mathrm { x } } \tan ^ { - 1 } \mathrm { t } ^ { 2 } \mathrm { dt } , [ 0,0.75 ]$$
A) $\frac { x ^ { 3 } } { 3 } - \frac { x ^ { 7 } } { 7 }$
B) $\frac { x ^ { 3 } } { 3 } - \frac { x ^ { 7 } } { 21 } + \frac { x ^ { 11 } } { 33 }$
C) $\frac { x ^ { 3 } } { 3 } - \frac { x ^ { 7 } } { 21 }$
D) $\frac { x ^ { 3 } } { 3 } - \frac { x ^ { 7 } } { 21 } + \frac { x ^ { 11 } } { 231 }$

Question

Quizplus · Accepted Answer

The polynomial of lowest degree that approximates $F(x)$ with an error less than $10^{-3}$ in the given interval is obtained by expanding the integrand $\tan^{-1}t^2$ using its Taylor series expansion and then integrating term by term. The Taylor series of $\tan^{-1}x$ is $x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots$. Substituting $x = t^2$, we get $t^2 - \frac{t^6}{3} + \frac{t^{10}}{5} - \frac{t^{14}}{7} + \cdots$. Integrating term by term from 0 to x gives $\frac{x^3}{3} - \frac{x^7}{21} + \frac{x^{11}}{55} - \frac{x^{15}}{105} + \cdots$. To ensure an error less than $10^{-3}$, we can stop at the term $\frac{x^7}{21}$, as higher-degree terms contribute significantly less to the value in the interval [0, 0.75], making option C the correct choice.

Find the Polynomial of Lowest Degree That Will Approximate F(x)F ( x )F(x)

Find the Polynomial of Lowest Degree That Will Approximate $F ( x )$