Practice of Evaluating Taylor Series

You may come across an integral that you cannot anti-differentiate through the usual integration techniques.

One such integral may be

Thankfully, a Taylor-Series Identity may be used to help solve the integral. As long as the limits of integration are within the interval of convergence of the series, then the value of the integral can be expressed as a Taylor Summation.

Note: The value received from the Taylor Summation will be correct only to a certain extent. Increasing the number of terms in the Taylor Summation will make the value more precise.

The technique can be seen in the works of the next example.

Example 7.3.1

Use the Taylor Series for ex to solve the integral

Example 7.3.2

Using the Taylor Series for cos x, evaluate the integral

Example 7.3.3

Find the anti-derivative of the Taylor Expansion of

Chapter 7 Problems

For questions 1-3, determine the value of the expansion

For questions 4 and 5, evaluate the infinite sum

The average liger weighs 6.2lbs at birth. One standard deviation is approximately 1.2lbs.

6. What is the probability of a liger weighing between 5 and 6.2 lbs?
7. What is the probability of a liger weighing between 6.8 and 7.4 lbs?
8. What is the probability of a liger weighing between 3.8 and 8.6 lbs?

For the next five questions, evaluate the definite integrals using a five-term Taylor Expansion

=

The expansion follows the series..

,

where x= . Therefore,

The expansion follows the series...

ex ≈

where x=420. Therefore, the expansion is equal to

e420

=

The expansion follows the series...

,

where . Therefore,

The summation follows the series..

Where x=1. Therefore,

sin-1(1) = .

The summation follows the alternating harmonic series. Expanding the summation...

The expansion follows the series...

where x=1. Therefore,

ln(2)=.693

6. A weight between 5 and 6.2 pounds falls one standard deviation from the norm.

The values for a and b are

a= -1, b=0. Plugging it into the formula..

. Using the exponential taylor series..

=34.1%

7. A weight between 6.8 and 7.4 pounds falls between half and one standard deviation from the norm.

The values for a and b are

a= .5, b=1. Plugging it into the formula..

. Using the exponential taylor series..

=14.99%

8. A weight between 3.8 and 8.6 pounds falls between two standard deviations to the left and right of the norm.

The values for a and b are

a= -2, b=2. Plugging it into the formula..

. Using the exponential taylor series..

$\frac{1}{\sqrt{2\pi }}[z-\frac{z^3}{6}+\frac{z^5}{40}-\frac{z^7}{336}+\frac{z^9}{3456}{]}_{-2}^2=\frac{2332}{945\sqrt{2\pi }}$

Using a five-term expansion..

=

= .998

Using a five-term expansion..

$\frac{x^2}{1}-\frac{x^4}{3!}+\frac{x^6}{5!}-\frac{x^8}{7!}+\frac{x^{10}}{9!}$

$\frac{x^3}{3}-\frac{x^5}{5\cdot 3!}+\frac{x^7}{7\cdot 5!}-\frac{x^9}{9\cdot 7!}+\frac{x^{11}}{11\cdot 9!}$

= $\frac{1^3}{3}-\frac{1^5}{5\cdot 3!}+\frac{1^7}{7\cdot 5!}-\frac{1^9}{9\cdot 7!}+\frac{1^{11}}{11\cdot 9!}$

= .301168

Using a five-term expansion..

`= 1.656

Using a five-term expansion..

$x^3/3-x^4/8+x^5/15-x^6/24+x^7/35$

= $(1/2{)}^3/3-(1/2{)}^4/8+(1/2{)}^5/15-(1/2{)}^6/24+(1/2{)}^7/35$

= .0355

Using a five-term expansion..

=

=

= .2496