## Taylor Polynomial Approximation of log(2.25)

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)

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.

2. Relevant 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$$\frac{\partial ^2 f}{\partial y ^2}$$ ]

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
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 Quote by 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: 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. 2. Relevant 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$$\frac{\partial ^2 f}{\partial y ^2}$$ ] 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
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 =)

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## Taylor Polynomial Approximation of log(2.25)

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).

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 Quote by RyanV 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.

 Quote by 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.
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 =)

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