# Taylor Polynomial Approximation of log(2.25)

• RyanV
In summary, you are trying to find a quadratic approximation to ln(2.25). You use a calculator to find the value of ln(2.25) to 4 decimal places, and comment on your answer.

## Homework Statement

Determine the order two Taylor polynomial, p2(x, y), for

f(x, y) = log e (1 + x2 + y4)

loge (2) + 2y - 2 + $$\frac{1}{2}$$ [ x2 - 2y2 + 4y - 2 ]

Managed that question and should be correct. If not, do let me know =)

Part 2: Using p2 (x, y), find a quadratic approximation to log e (2.25) to 4 decimal places. Use a calculator to find the value of log e (2.25) to 4 decimal places, and comment on your answer.

## Homework Equations

2nd Order Taylor Polynomial
p2(x, y) = f(x, y) + (x + a)$$\frac{\partial f}{\partial x}$$ + (y + b)$$\frac{\partial f}{\partial y}$$ + $$\frac{1}{2}$$ [ (x - a)2$$\frac{\partial ^2 f}{\partial x^2}$$ + (x - a)(y - b)$$\frac{\partial ^2 f}{\partial x\partial y}$$ + (y - b)2$$\frac{\partial ^2 f}{\partial y ^2}$$ ]

## The Attempt at a Solution

I'm not too sure where to go from there...I thought of putting p2(x, y) as log 2 (2.25) but I'm not sure what that does after simplifying...

Any assistance would be of much help! =)
Thanks

RyanV said:

## Homework Statement

Determine the order two Taylor polynomial, p2(x, y), for

f(x, y) = log e (1 + x2 + y4)

loge (2) + 2y - 2 + $$\frac{1}{2}$$ [ x2 - 2y2 + 4y - 2 ]

Managed that question and should be correct. If not, do let me know =)
You got the sign of fyy wrong.
Part 2: Using p2 (x, y), find a quadratic approximation to log e (2.25) to 4 decimal places. Use a calculator to find the value of log e (2.25) to 4 decimal places, and comment on your answer.

## Homework Equations

2nd Order Taylor Polynomial
p2(x, y) = f(x, y) + (x + a)$$\frac{\partial f}{\partial x}$$ + (y + b)$$\frac{\partial f}{\partial y}$$ + $$\frac{1}{2}$$ [ (x - a)2$$\frac{\partial ^2 f}{\partial x^2}$$ + (x - a)(y - b)$$\frac{\partial ^2 f}{\partial x\partial y}$$ + (y - b)2$$\frac{\partial ^2 f}{\partial y ^2}$$ ]

## The Attempt at a Solution

I'm not too sure where to go from there...I thought of putting p2(x, y) as log 2 (2.25) but I'm not sure what that does after simplifying...

Any assistance would be of much help! =)
Thanks
What do you mean by "putting p2(x, y) as log 2 (2.25)"?

Just choose values of x and y so that f(x,y)=log 2.25 and plug those values into your expansion.

I got fyy = -2

This was from fyy = $$\frac{12y^2( 1 + x^2 + y^4 ) - 16y^2}{(1 + x^2 + y^4 ) ^2}$$

and since at point (0,1),

fyy = $$\frac{12 (1)^2( 1 + (0)^2 + (1)^4 ) - 16 (1)^2}{(1 + (0)^2 + (1)^4 ) ^2}$$
fyy = -2

Isn't it? Unless I stuffed up the Taylor polynomial part somewhere...

Oh, when I meant putting p2(x, y) as log e (2.25) [correction], I meant like:

loge(2.25) = loge (2) + 2y - 2 + $$\frac{1}{2}$$ [ x2 - 2y2 + 4y - 2 ]

But yeah..got me no where =(

I don't understand how I can choose values for x and y because how can we figure out what is loge (2.25) from something like loge (2) + constant. Seeing that we just choose values for x and y...unless there's the use of a calculator. Could you help shed some light on this? Thanks =)

You have f(x, y) = ln (1 + x2 + y4),
so f(0, 1) = ln(1 + 02 + 12) = ln 2

You want ln 2.25, which can be represented as f(.5, 1). Use the polynomial you found to approximate f(.5, 1) by p2(.5, 1).

RyanV said:
I got fyy = -2

This was from fyy = $$\frac{12y^2( 1 + x^2 + y^4 ) - 16y^2}{(1 + x^2 + y^4 ) ^2}$$

and since at point (0,1),

fyy = $$\frac{12 (1)^2( 1 + (0)^2 + (1)^4 ) - 16 (1)^2}{(1 + (0)^2 + (1)^4 ) ^2}$$
fyy = -2

Isn't it? Unless I stuffed up the Taylor polynomial part somewhere...
No, the top is 12x2-16 = 24-16 = 8, and the bottom is 4.
Oh, when I meant putting p2(x, y) as log e (2.25) [correction], I meant like:
The base of the log didn't confuse me; I figured that was a typo. It was your wording, particularly your choice of verb.

vela said:
No, the top is 12x2-16 = 24-16 = 8, and the bottom is 4.

The base of the log didn't confuse me; I figured that was a typo. It was your wording, particularly your choice of verb.

Oh sorry. What would have been a better choice of verb?

Thanks a lot guys, I believe I managed to get it now.

Worked it out to be loge (2) + $$\frac{1}{8}$$ = 0.8181

Hopefully that's right =)

## What is Taylor Polynomial Approximation?

Taylor Polynomial Approximation is a mathematical method used to approximate a function using a polynomial expression. It involves choosing a point of approximation and calculating the derivatives of the function at that point to create a polynomial expression that closely resembles the original function.

## How does Taylor Polynomial Approximation work?

Taylor Polynomial Approximation works by using derivatives of a function at a chosen point to create a polynomial expression that approximates the function. The more derivatives that are used, the closer the polynomial will be to the original function. This method is based on the Taylor Series expansion, which can be used to approximate a function at any point.

## What is the purpose of using Taylor Polynomial Approximation?

The purpose of using Taylor Polynomial Approximation is to estimate the value of a function at a specific point or to approximate the behavior of a function near a given point. This method is particularly useful when the exact value of a function cannot be calculated or when it is difficult to graph the function.

## How is Taylor Polynomial Approximation used in the calculation of log(2.25)?

In the calculation of log(2.25), Taylor Polynomial Approximation can be used to approximate the natural logarithm function. By choosing a point of approximation (such as x = 1), the derivatives of the natural logarithm function can be calculated and used to create a polynomial expression that approximates the function. This polynomial can then be used to calculate the value of log(2.25) with a reasonable degree of accuracy.

## What are the advantages of using Taylor Polynomial Approximation?

One advantage of using Taylor Polynomial Approximation is that it allows us to approximate the behavior of a function near a given point without having to know the exact value of the function at that point. It can also be used to approximate functions that are difficult to graph or calculate. Additionally, by using more derivatives in the polynomial expression, we can achieve a more accurate approximation of the function.