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

In summary, the conversation discusses finding the Taylor polynomial of order 3 at (0,0,0) for the function f(x,y,z)=\sqrt{e^{-x}+\sin y+z^{2}} using well-known series, rather than taking derivatives directly. The suggested approach is to write the function as a Taylor polynomial and then remove the square root. However, there is some confusion about how to properly remove the square root and calculate the Taylor series at x=0.
  • #1
gop
58
0

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)

[tex]f(x,y,z)=\sqrt{e^{-x}+\sin y+z^{2}}[/tex]

Homework Equations





The Attempt at a Solution



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

[tex]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}}[/tex]

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 [tex]\sqrt{1+x}[/tex] but that didn't really work out very well.
 
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  • #2
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?
 
  • #3
Reference The formula for the Taylor series expansion of [itex]f(x,y,z)[/itex] about the point [itex](x_0,y_0,z_0)[/itex] is

[tex]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!} [/tex]​

Use To compute the Tayor polynomial of order 3, only write out the terms for which [itex]n+k+j\le 3[/itex].
 
  • #4
As stated I have to use "well-known series" to arrive at the taylor polynomial; thus, I'm not allowed to just take derivatives.
 
  • #5
Sorry, I missed reading that part!

Okay, what is the Taylor's series for [iitex]\sqrt{x}[/itex]?
 
  • #6
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.
 

What is a Taylor Polynomial of Order 3?

A Taylor Polynomial of Order 3 is a mathematical approximation of a function using a polynomial of degree 3. It is used to estimate the value of the function at a specific point by using the function's values and derivatives at that point.

What is the purpose of finding the Taylor Polynomial of Order 3 for a function at (0,0,0)?

The Taylor Polynomial of Order 3 at (0,0,0) is used to approximate the function at the origin. This can be helpful in situations where the function is difficult to evaluate at that point, or when an exact value is not necessary.

How is the Taylor Polynomial of Order 3 calculated?

The Taylor Polynomial of Order 3 is calculated using the function's values and derivatives at the point of interest. The formula for the Taylor Polynomial of Order 3 is f(x,y,z) = f(0,0,0) + fx(0,0,0)x + fy(0,0,0)y + fz(0,0,0)z + (1/2!)(fxx(0,0,0)x2 + fyy(0,0,0)y2 + fzz(0,0,0)z2) + (1/3!)(fxxx(0,0,0)x3 + fyyy(0,0,0)y3 + fzzz(0,0,0)z3)

What is the difference between a Taylor Polynomial of Order 3 and a Taylor Series?

A Taylor Polynomial of Order 3 is a polynomial of degree 3 used to approximate a function at a specific point, while a Taylor Series is an infinite sum of polynomials used to represent a function at all points within a certain interval. A Taylor Polynomial is a finite approximation, while a Taylor Series is an infinite representation of the function.

What is the significance of the Taylor Polynomial of Order 3 in calculus?

The Taylor Polynomial of Order 3 is important in calculus because it is used to approximate functions and make calculations easier. It is also used in the study of convergence and divergence of series, and in the development of more complex mathematical concepts such as Taylor's theorem and Taylor series.

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