Taylor Polynomial Approximation of log(2.25)

RyanV

1. The problem statement, all variables and given/known data
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 + [tex]\frac{1}{2}[/tex] [ 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.


2. Relevant equations

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


3. 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

You got the sign of f_yy wrong.
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 = [tex]\frac{12y^2( 1 + x^2 + y^4 ) - 16y^2}{(1 + x^2 + y^4 ) ^2}[/tex]

and since at point (0,1),

f_yy = [tex]\frac{12 (1)^2( 1 + (0)^2 + (1)^4 ) - 16 (1)^2}{(1 + (0)^2 + (1)^4 ) ^2}[/tex]
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 + [tex]\frac{1}{2}[/tex] [ 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

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.

RyanV

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) + [tex]\frac{1}{8}[/tex] = 0.8181

Hopefully that's right =)