Taylor Polynomial Approximation of log(2.25)

[RyanV]

Homework Statement

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

f(x, y) = log _e (1 + x² + y⁴)

about point (0, 1)


ANSWER:
log_e (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

 

[vela]

Homework Statement

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

f(x, y) = log _e (1 + x² + y⁴)

about point (0, 1)


ANSWER:
log_e (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 f_yy 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.

 

[RyanV]

I got f_yy = -2

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

and since at point (0,1),

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

log_e(2.25) = log_e (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 log_e (2.25) from something like log_e (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 =)

 

[Mark44]

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).

 

[vela]

I got f_yy = -2

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

and since at point (0,1),

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

 

[RyanV]

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 log_e (2) + $$\frac{1}{8}$$ = 0.8181

Hopefully that's right =)