Taylor Polynomial of Order 3 for f(x,y,z) at (0,0,0)

[gop]

Homework Statement

Calculate the taylor polynom of order 3 at (0,0,0) of the function with well-known series (that means I can't just take the derivatives)

$$f(x,y,z)=\sqrt{e^{-x}+\sin y+z^{2}}$$

Homework Equations

The Attempt at a Solution

I wrote the functions within the square root as taylor polynomials and got

$$f(x,y,z)=\sqrt{1+-x+\frac{1}{2}x^{2}-\frac{1}{6}x^{3}+y-\frac{1}{6}y^{3}+z^{2}}$$

But then I don't really know how to "remove" the square root. I already tried to just plug the term inside the square root in the taylor expansion of $$\sqrt{1+x}$$ but that didn't really work out very well.

 

[HallsofIvy]

Why did you do that? What not just write [itex]f(x,y,z)= (e^{-x}+ sin(y)+ z^2)^{1/2}[itex] and calculate the derivatives?

 

[benorin]

Reference The formula for the Taylor series expansion of $f(x,y,z)$ about the point $(x_0,y_0,z_0)$ is

$$f(x,y,z)=\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\sum_{j=0}^{\infty} \frac{\partial ^{n+k+j}f (x_0,y_0,z_0)}{\partial x^{n}\partial y^{k}\partial z^{j}} \cdot\frac{(x-x_0)^{n}}{n!} \cdot\frac{(y-y_0)^{k}}{k!}\cdot\frac{(z-z_0)^{j}}{j!}$$​

Use To compute the Tayor polynomial of order 3, only write out the terms for which $n+k+j\le 3$.

 

[gop]

As stated I have to use "well-known series" to arrive at the taylor polynomial; thus, I'm not allowed to just take derivatives.

 

[HallsofIvy]

Sorry, I missed reading that part!

Okay, what is the Taylor's series for [iitex]\sqrt{x}[/itex]?

 

[gop]

I can't really calculate the taylor series at x=0 because the derivatives is then of the form 1/0 and doesn't exist. I already tried sqrt(1+x) but that didn't produce a correct result.