# Third degree Taylor polynomial in two variables

1. Oct 24, 2012

### 5hassay

1. The problem statement, all variables and given/known data

Find the third degree Taylor polynomial about the origin of

$f(x,y) = \frac{\cos(x)}{1+xy}$

2. Relevant equations

3. The attempt at a solution

From my ventures on the Internet, this is my attempt:

I see that

$\cos(x) = 1 - \frac{1}{2}x^2 + \frac{1}{4!}x^4 - \cdots$

$\frac{1}{1+x} = 1 - x + x^2 - x^3 + \cdots$

and so

$\frac{1}{1+(xy)} = 1 - xy + x^2y^2 - x^3y^3 + \cdots$

Therefore, in multiplying them out,

$f(x,y) = \frac{\cos(x)}{1+xy} = 1 - xy + x^2y^2 - x^3y^3 - x + x^2y - x^3y^2 + \cdots$

And I suppose that would be my answer.

Do I have the right idea?

Thanks in advance.

EDIT: Oops! I didn't substitute $(x,y) = (0,0)$. So, in doing that, I should get precisely $1$. That is my answer.

EDIT: But $1$ doesn't seem right...

EDIT: I think I am getting confused. Pretty sure the product above would be the answer, D=

Last edited: Oct 24, 2012
2. Oct 24, 2012

### Zondrina

Okay so, first things first, recall the taylor series of cos(x) around a = 0 ( Aka the maclaurin series ). As well as the one for your other fraction there which I believe you've already done.

Now you want at most a polynomial of degree three, correct? So take all your terms from both series which are polynomials of degree three or less, so for example for cosx you would choose :

$1 - \frac{1}{2} x^2$

Now do the same for your other series and multiply the two resulting equations together. What do you get?

3. Oct 24, 2012

### 5hassay

Thanks for the reply, Zondrina.

Ohhh, so that is what is meant by "$n$-th degree!"

Alright, then I would have

$f(x,y) = (1 - \frac{1}{2}x^2) (1 - xy + x^2y^2 - x^3y^3)$

so

$f(x,y) = 1 - xy + x^2y^2 -x^3y^3 - \frac{1}{2}x^2 + \frac{1}{2}x^3y - \frac{1}{2}x^4y^2 + \frac{1}{2}x^5y^3$

However, as you said, I only want degree three or less, so I suppose I should eliminate all terms consisting of variables of degree greater than three, getting

$f(x,y) = 1 - xy + x^2y^2 -x^3y^3 - \frac{1}{2}x^2 + \frac{1}{2}x^3y$

And, this would be my answer?

4. Oct 24, 2012

### HallsofIvy

Staff Emeritus
You are still mis-understanding what "nth degree" meas. $x^2y^2$ is of degree 4 and $x^3y^3$ is of degree 6.

5. Oct 24, 2012

### 5hassay

Thanks for the reply, HallsofIvy.

Oh dear, I see. Hopefully this will be my last change of answer:

$f(x,y) = (1 - \frac{1}{2}x^2) (1 - xy) = 1 - xy - \frac{1}{2}x^2$

Do I now have the correct idea? D=

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