Third degree Taylor polynomial in two variables

[5hassay]

Homework Statement

Find the third degree Taylor polynomial about the origin of

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

Homework Equations

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=

 

[STEMucator]

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?

 

[5hassay]

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?

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?

 

[HallsofIvy]

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?

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

 

[5hassay]

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

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=