# Taylor Polynomial of degree 2

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• August 12th 2010, 03:02 AM
tsang
Taylor Polynomial of degree 2
Hi friends, I'm having trouble with this Taylor Polynomial question.

f(x,y)=arctan(x/y)
Using Taylor Polynomial degree 2, find a quadratic approximation to arctan(1/3) to 4 decimal places. Use a calculator to find the value of arctan(1/3) to 4 decimal places, and comment on your answers.

I can use formula to get Taylor Polynomial to degree 2 for f(x,y), but I supposed to do next? Normally the question should give something like (x,y)=(1.98,3.02) etc, so we would find a close integer and apply Taylor Polynomial. In this case, I have no idea what I should do.
Also, what does the question mean by comment on my answers? Does that mean compare my answer with real solution?
Thanks.
• August 12th 2010, 03:23 AM
Vlasev
When you are constructing the Taylor polynomial (of any degree) you need to construct it around a point. Have you been given one such point? Usually, for hand calculations, you want to pick a point that is relatively close to the point you are trying to calculate. At the same time you want the Taylor polynomial to be easy to calculate. Have you been given a point to construct around? If not, you need to find the value of \arctan(1/3), which corresponds to x = 1 and y = 3. You cannot center the polynomial around this point, so you have to center it at a point that is close by and makes calculation easy.

The partial derivatives of this function are just fractions, so it doesn't really matter which point you choose. Which point did you choose? There are two that may interest you: (0,1) and (1,2) which are both the same distance. The first one has the added advantage that you are setting all x's to 0 which makes the form really simple. Furthermore, for the degree 0 term you will have to calculate arctan(0/3) which is just 0.

Of course, if you are more concerned about the accuracy you can just pick the point (1.98,3.02) as your center, but that will require you to calculate arctan(1.98/3.02) to a good precision for your constant term, which defeats the pupose. Now, commenting your work really means to just compare the answer you got to the actual answer and maybe say what percentage it is off by.
• August 12th 2010, 04:29 AM
tsang
Quote:

Originally Posted by Vlasev
When you are constructing the Taylor polynomial (of any degree) you need to construct it around a point. Have you been given one such point? Usually, for hand calculations, you want to pick a point that is relatively close to the point you are trying to calculate. At the same time you want the Taylor polynomial to be easy to calculate. Have you been given a point to construct around? If not, you need to find the value of \arctan(1/3), which corresponds to x = 1 and y = 3. You cannot center the polynomial around this point, so you have to center it at a point that is close by and makes calculation easy.

The partial derivatives of this function are just fractions, so it doesn't really matter which point you choose. Which point did you choose? There are two that may interest you: (0,1) and (1,2) which are both the same distance. The first one has the added advantage that you are setting all x's to 0 which makes the form really simple. Furthermore, for the degree 0 term you will have to calculate arctan(0/3) which is just 0.

Of course, if you are more concerned about the accuracy you can just pick the point (1.98,3.02) as your center, but that will require you to calculate arctan(1.98/3.02) to a good precision for your constant term, which defeats the pupose. Now, commenting your work really means to just compare the answer you got to the actual answer and maybe say what percentage it is off by.

Dear Vlasev, firstly, thank you so much for your help. I know you have helped me so many times already, and I really appreciate, thank you.

And secondly, yes, this question was made by three parts, the first part was asking to determine the Taylor Polynomial of degree 2 for arctan(x/y) about (x,y)=(0,1), and the second part was check the the value of all partial derivatives when (x,y)=(0,1).

Then, the third part is the one that I had above and struggling with.
I did try to use the point (x,y)=(0,1), but I end up with answer 3, which certainly looks not right, and I certainly cannot get up to 4 decimals, unless I write it as 3.0000.
I used the calculator to find the answer should be 0.3217. Then how can I have my answer is 3? Could you please help me with it? Thanks a lot.
• August 12th 2010, 05:37 AM
tsang
I've just tried to use point (0,1) again, and it still doesn't work, I double checked that I do end up with answer 3, which is too far from 0.3217, what should I do really?
• August 12th 2010, 06:15 AM
Jester
Not sure how you got 3. If you post for work, then amybe we can comment. I centered about (0,2) for the ease of calculating the first term in the seris (arctan 0 = 0). What I ended up with was

$\arctan\dfrac{x}{y} \approx P_2(x,y) = \dfrac{x}{2} - \dfrac{x(y-2)}{4}$

Then

$P_2(1,3) = \dfrac{1}{4} = 0.2500$

$\arctan(1/3) = 0.3218$

You can see the answer really isn't that good! The absolute error is E = 0.0718.

I think the question was asking for your answer to 4 decimal place not having the approximate answer accurate to 4 decimal places.
• August 13th 2010, 12:37 AM
Ares_D1
Quote:

Originally Posted by Danny
Not sure how you got 3. If you post for work, then amybe we can comment. I centered about (0,2) for the ease of calculating the first term in the seris (arctan 0 = 0). What I ended up with was

$\arctan\dfrac{x}{y} \approx P_2(x,y) = \dfrac{x}{2} - \dfrac{x(y-2)}{4}$

Then

$P_2(1,3) = \dfrac{1}{4} = 0.2500$

$\arctan(1/3) = 0.3218$

You can see the answer really isn't that good! The absolute error is E = 0.0718.

I think the question was asking for your answer to 4 decimal place not having the approximate answer accurate to 4 decimal places.

I believe the question has asked us to center around (0,1), although it doesn't say explicitly. But using this I end up with
$\arctan\dfrac{x}{y} \approx P_2(x,y) = x(2-y)$

So
$P_2(1,3) = -1$

Obviously not really the answer I was looking for. And I have checked my work very thoroughly - can not identify any errors.
• August 15th 2010, 03:51 PM
nugunderground
Quote:

Originally Posted by Ares_D1
I believe the question has asked us to center around (0,1), although it doesn't say explicitly. But using this I end up with
$\arctan\dfrac{x}{y} \approx P_2(x,y) = x(2-y)$

So
$P_2(1,3) = -1$

Obviously not really the answer I was looking for. And I have checked my work very thoroughly - can not identify any errors.

I just did it myself and got the same as you, however I'm not entirely sure if this is correct either... (Thinking)