# Taylor Polynomials And Approximations Homework

## How to Use Taylor Polynomials and Approximations to Solve Homework Problems

Taylor polynomials and approximations are powerful tools that can help you solve various homework problems in calculus, physics, engineering, and other fields. They allow you to approximate complicated functions with simpler ones that are easier to work with. In this article, we will explain what Taylor polynomials and approximations are, how to find them, and how to use them to solve homework problems.

## Taylor Polynomials And Approximations Homework

## What are Taylor Polynomials and Approximations?

A Taylor polynomial is a polynomial function that approximates another function near a given point. A Taylor approximation is the value of a Taylor polynomial at a given point. For example, suppose you have a function f(x) and you want to approximate it near x = a. You can find a Taylor polynomial P(x) of degree n such that P(a) = f(a), P'(a) = f'(a), P''(a) = f''(a), ..., P^(n)(a) = f^(n)(a), where P^(n) denotes the nth derivative of P. The value of P(x) at x = a is called the Taylor approximation of f(x) at x = a.

The idea behind Taylor polynomials and approximations is that if two functions have the same value and the same derivatives up to a certain order at a point, then they are very close to each other near that point. The higher the degree of the Taylor polynomial, the better the approximation. However, finding higher degree Taylor polynomials can be tedious and time-consuming, so sometimes it is enough to use lower degree ones for homework purposes.

## How to Find Taylor Polynomials and Approximations?

To find a Taylor polynomial of degree n for a function f(x) near x = a, you need to use the following formula:

P(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2! + ... + f^(n)(a)(x-a)^n/n!

This formula is called the Taylor series expansion of f(x) at x = a. It is an infinite series that converges to f(x) as n goes to infinity. However, for practical purposes, we only use the first few terms of the series as our Taylor polynomial.

To find a Taylor approximation of f(x) at x = b, you simply need to plug in x = b into the Taylor polynomial P(x). For example, suppose you want to find a Taylor approximation of sin(x) at x = pi/6 using a Taylor polynomial of degree 3 near x = 0. You can use the following steps:

Find the first four derivatives of sin(x): sin'(x) = cos(x), sin''(x) = -sin(x), sin'''(x) = -cos(x), sin''''(x) = sin(x).

Evaluate these derivatives at x = 0: sin(0) = 0, cos(0) = 1, -sin(0) = 0, -cos(0) = -1.

Plug these values into the Taylor series formula: P(x) = 0 + 1(x-0) + 0(x-0)^2/2! + -1(x-0)^3/3!.

Simplify the expression: P(x) = x - x^3/6.

Plug in x = pi/6 into P(x): P(pi/6) = pi/6 - (pi/6)^3/6.

Calculate the result: P(pi/6) 0.5 - 0.0073 0.4927.

This is the Taylor approximation of sin(pi/6) using a Taylor polynomial of degree 3 near x = 0.

## How to Use Taylor Polynomials and Approximations to Solve Homework Problems?

One of the main applications of Taylor polynomials and approximations is to solve homework problems that involve complicated functions that are hard to manipulate or evaluate. By using Taylor polynomials and approximations, we can replace these functions with simpler ones that are easier to work with. Here are some examples of how to use Taylor polynomials and approximations to solve homework problems:

### Example 1: Evaluating Limits

Suppose we want to evaluate the limit:

lim x0 (1 cosx) / x2

This limit is indeterminate of the form 0 / 0, so we cannot use direct substitution. However, we can use a Taylor polynomial of degree 2 for cosx near x = 0 to approximate it:

cosx 1 x2 / 2

This approximation comes from the Taylor series expansion of cosx at x = 0, which is:

cosx = 1 x2 / 2! + x4 / 4! x6 / 6! + ...

We only use the first two terms as our Taylor polynomial of degree 2. Now we can plug this approximation into the limit and simplify:

lim x0 (1 cosx) / x2 lim x0 (1 (1 x2 / 2)) / x2

= lim x0 (x2 / 2) / x2

= lim x0 1 / 2

= 1 / 2

Therefore, the limit is approximately 1 / 2.

### Example 2: Approximating Definite Integrals

Suppose we want to approximate the definite integral:

0.5 ex^2 dx

This integral cannot be evaluated exactly using elementary functions, so we need to use numerical methods or approximation techniques. One way to approximate it is to use a Taylor polynomial of degree 6 for ex^2 near x = 0:

ex^2 1 x^2 + x^4 / 2 x^6 / 6

This approximation comes from the Taylor series expansion of ex^2 at x = 0, which is:

ex^2 = 1 x^2 + x^4 / 2! x^6 / 6! + ...

We only use the first four terms as our Taylor polynomial of degree 6. Now we can plug this approximation into the integral and evaluate it:

0.5 ex^2 dx 0.5 (1 x^2 + x^4 / 2 x^6 / 6) dx

= [x x3 / 3 + x5 / 10 x7 / 42] 0.5

= (0.5 (0.5)^3 / 3 + (0.5)^5 / 10 (0.5)^7 / 42) (0 0^3 / 3 + 0^5 / 10 0^7 /42)

= (0.5 (1/8) /3 + (1/32) /10 - (1/128) /42) - (0)

= (20/32 - (4/32) +(3/32) - (3/32)) - (0)

= (16/32) - (0)

= (1/2) - (0)

= (1/2)

### Example 3: Finding Maclaurin Series

A Maclaurin series is a special case of a Taylor series where the center point is x = 0. A Maclaurin series can be used to represent any function that is infinitely differentiable at x = 0. To find a Maclaurin series for a function f(x), we need to find the Taylor polynomial of degree n for f(x) near x = 0 and then let n go to infinity. The general formula for a Maclaurin series is:

f(x) = f(0) + f'(0)x + f''(0)x^2 / 2! + ... + f^(n)(0)x^n / n! + ...

For example, suppose we want to find the Maclaurin series for sinx. We can use the following steps:

Find the derivatives of sinx: sin'(x) = cosx, sin''(x) = -sinx, sin'''(x) = -cosx, sin''''(x) = sinx, and so on.

Evaluate these derivatives at x = 0: sin(0) = 0, cos(0) = 1, -sin(0) = 0, -cos(0) = -1, and so on.

Plug these values into the Maclaurin series formula: sinx = 0 + 1x + 0x^2 / 2! + -1x^3 / 3! + ...

Simplify the expression: sinx = x - x^3 / 6 + x^5 / 120 - x^7 / 5040 + ...

This is the Maclaurin series for sinx.

### Example 4: Finding Taylor Series

A Taylor series is a generalization of a Maclaurin series where the center point can be any value of x. A Taylor series can be used to represent any function that is infinitely differentiable at some point x = a. To find a Taylor series for a function f(x), we need to find the Taylor polynomial of degree n for f(x) near x = a and then let n go to infinity. The general formula for a Taylor series is:

f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2 / 2! + ... + f^(n)(a)(x-a)^n / n! + ...

For example, suppose we want to find the Taylor series for ex centered at x = 1. We can use the following steps:

Find the derivatives of ex: e'x = ex, e''x = ex, e'''x = ex, e''''x = ex, and so on.

Evaluate these derivatives at x = 1: e^1 = e, e^1 = e, e^1 = e, e^1 = e, and so on.

Plug these values into the Taylor series formula: ex = e + e(x-1) + e(x-1)^2 / 2! + ...

Simplify the expression: ex = e(1 + (x-1) + (x-1)^2 / 2! + ...)

This is the Taylor series for ex centered at x = 1.

### Example 5: Finding Radius and Interval of Convergence

A Taylor series is not always valid for all values of x. Sometimes, it only converges to the original function for x within a certain range. The radius of convergence of a Taylor series is the distance from the center point x = a to the nearest point where the series diverges. The interval of convergence of a Taylor series is the set of all x values for which the series converges. To find the radius and interval of convergence of a Taylor series, we need to use some tests for convergence of infinite series, such as the ratio test or the root test. For example, suppose we want to find the radius and interval of convergence of the Taylor series for 1 / (1 + x^2) centered at x = 0:

1 / (1 + x^2) = 1 x^2 + x^4 x^6 + ...

We can use the ratio test to determine the radius of convergence. The ratio test states that if lim n an + 1 / an = L, then the series an converges if L 1. In this case, we have:

an = (1)n x2n

an + 1 = (1)n + 1 x2n + 2

an + 1 / an = (1)n + 1 x2n + 2 / (1)n x2n = x^2

lim n an + 1 / an = lim n x^2 = x^2

This limit is less than 1 when x^2 1. Therefore, the radius of convergence is R = 1.

To find the interval of convergence, we need to check the endpoints of the interval x

n=0 (1)n (1)2n = n=0 (1)3n = n=0 (1)n

This is an alternating series that does not converge, since its terms do not approach zero as n goes to infinity. When x = 1, we have:

n=0 (1)n (1)2n = n=0 (1)n

This is also an alternating series that does not converge, for the same reason as before. Therefore, neither endpoint is included in the interval of convergence. The interval of convergence is (1, 1).

### Example 6: Finding Error Bounds

When we use a Taylor polynomial to approximate a function, we need to know how accurate our approximation is. The error bound of a Taylor polynomial is an estimate of the maximum possible difference between the original function and the polynomial for x values within a certain range. One way to find an error bound is to use the Lagrange error formula, which states that if f(x) has n + 1 continuous derivatives on an interval containing x = a and x = b, then there exists some z between a and b such that:

f(x) pn(x) = f^(n + 1)(z) (x a)n + 1 / (n + 1)!

where pn(x) is the Taylor polynomial of degree n for f(x) centered at x = a. The absolute value of the right-hand side of this formula gives us an upper bound for the error f(x) pn(x) . To use this formula, we need to find an upper bound for f^(n + 1)(z) on the interval of interest. For example, suppose we want to find an error bound for using p3(x) = x x^3 / 6 to approximate sinx on [0, 0.5]. We can use the following steps:

Find the fourth derivative of sinx: sin'(x) = cosx, sin''(x) = -sinx, sin'''(x) = -cosx, sin''''(x) = sinx.

Find an upper bound for sin^(4)(z) on [0, 0.5]: Since sinx is bounded by -1 and 1 for all x, we have sin^(4)(z) 1 for all z on [0, 0.5].

Plug these values into the Lagrange error formula: sinx p3(x) sin^(4)(z) (x 0)^4 / 4! (1)(x)^4 / 24 .

Simplify the expression: sinx p3(x) x^4 / 24.

This means that the maximum possible difference between sinx and p3(x) on [0, 0.5] is x^4 / 24.

## Conclusion

In this article, we have learned how to use Taylor polynomials and approximations to solve homework problems involving complicated functions. We have seen how to find Taylor polynomials of different degrees for different functions centered at different points. We have also seen how to use Taylor approximations to evaluate limits, approximate definite integrals, find Maclaurin and Taylor series, find radius and interval of convergence, and find error bounds. We have learned that Taylor polynomials and approximations are powerful tools that can help us simplify and understand complex functions. However, we have also learned that they have some limitations and challenges, such as choosing the appropriate degree and center point, finding the derivatives and values of the function, applying the tests for convergence, and estimating the error. Therefore, we need to use Taylor polynomials and approximations with care and caution, and always check our results for accuracy and validity. d282676c82

__https://www.sofyortiz.coach/group/mysite-200-group/discussion/16f87126-3333-4a22-a547-da5bc3596a44__