Power series of a function of 2 variables

In summary, the Taylor series for a function of two real variables can be derived in a similar way to that for a function of one.
  • #1
Castilla
241
0
I have learned that if a function of one real variable can be defined as a power series, then this one is its Taylor series.

Does the same occur with functions of 2 real variables? I mean, if a function f(x, y) can be defined as a power series, does this series is the Taylor series of f(x, y)?

Thanks for help.
 
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  • #2
I thought an expansion like this was possible...
[tex]
f(x, y) = f(x_0, y_0)
+ \left. \frac{\partial f(x, y_0)}{\partial x} \right|_{x = x_0} (x - x_0)
+ \left. \frac{\partial f(x_0, y)}{\partial x} \right|_{y = y_0} (y - y_0)
+ \frac12 \left. \frac{\partial^2 f(x, y_0)}{\partial x^2} \right|_{x = x_0} (x - x_0)^2
[/tex][tex]
+ \frac12 \left. \frac{\partial^2 f(x_0, y)}{\partial y^2} \right|_{y = y_0} (y - y_0)^2
+ \frac12 \left. \frac{\partial^2 f(x, y)}{\partial x \partial y}(x-x_0)(y-y_0) \cdots
+ \mathcal{O}(x, y)^3
[/tex]
 
Last edited by a moderator:
  • #3
That should be
[tex]
f(x, y) = f(x_0, y_0)
+ \left. \frac{\partial f(x, y_0)}{\partial x} \right|_{x = x_0} (x - x_0)
+ \left. \frac{\partial f(x_0, y)}{\partial x} \right|_{y = y_0} (y - y_0)
+ \frac12 \left. \frac{\partial^2 f(x, y_0)}{\partial x^2} \right|_{x = x_0} (x - x_0)^2
[/tex][tex]
+ \frac12 \left. \frac{\partial^2 f(x_0, y)}{\partial y^2} \right|_{y = y_0} (y - y_0)^2
+ \frac12 \left. \frac{\partial^2 f(x, y)}{\partial x \partial y}(x-x_0)(y-y_0)\cdots
+ \mathcal{O}(x, y)^3
[/tex]
where I have added [itex](x-x_0)(y-y_0)[/itex] after the mixed second derivative.
 
  • #4
I believe that's what I meant by the [itex]\cdots[/itex], sorry for being unclear.
 
  • #5
This is incorrect, Halls!
You have a 1/2 in front of the mixed second partial; it should be a 1 instead.

For OP:
Here's how we can DEDUCE the look of the Taylor polynomial for a 2-variable function.
Now, given a function f(x,y); we may as a first step regard this as a single variable function G(x;y)=f(x,y); where "y" in G is just some fixed parameter.
G can be expanded in a 1-variable Taylor series in x about the point (x0,y), so switching to f-notation, we have:
[tex]f(x,y)=\sum_{n=0}^{\infty}\frac{1}{n!}\frac{\partial^{n}f}{\partial{x}^{n}}(x_{0},y)(x-x_{0})^{n}[/tex]
where the 0'th derivative of a function means the function itself.

Now, each of these derivatives is a function of y, with a fixed parameter x0. Thus, they can be expanded as Taylor series, and we get:
[tex]f(x,y)=\sum_{n=0}^{\infty}\frac{1}{n!}\sum_{m=0}^{\infty}\frac{1}{m!}\frac{\partial^{(n+m)}f}{\partial{x}^{n}\partial{y}^{m}}(x_{0},y_{0})(x-x_{0})^{n}(y-y_{0})^{m}[/tex]
Regrouping our double series in term of the total derivative index s=n+m, we readily get:
[tex]f(x,y)=\sum_{s=0}^{\infty}\frac{1}{s!}\sum_{n=0}^{s}\binom{s}{n}\frac{\partial^{s}f}{\partial{x}^{n}\partial{y}^{(s-n)}}(x_{0},y_{0})(x-x_{0})^{n}(y-y_{0})^{(s-n)}[/tex]
where I have utilized [tex]\frac{1}{s!}\binom{s}{n}=\frac{1}{n!}\frac{1}{(s-n)!}[/tex]


This form is readily extendable to functions with more than two variables as well.
 
Last edited:
  • #6
One easy way to remember the Taylor series in higher dimensions is to write it like

[tex]
f(x+u) = f(x)\; +\; u\cdot\nabla f(x)\; +\; \frac{1}{2!}(u\cdot\nabla)^2 f(x)\; +\; \frac{1}{3!}(u\cdot\nabla)^3 f(x)\; + \cdots
[/tex]

You can get those coefficients for the partial derivatives by computing [tex](u\cdot\nabla)^n[/tex] open. For example

[tex]
(u_1\partial_1 + u_2\partial_2)^2 = u_1^2\partial_1^2 + 2u_1u_2\partial_1\partial_2 + u_2^2\partial_2^2
[/tex]
 

1. What is a power series of a function of 2 variables?

A power series of a function of 2 variables is a mathematical expression that represents a function as an infinite sum of terms, where each term is a polynomial with increasing powers of the two variables.

2. How is a power series of a function of 2 variables different from a power series of a function of 1 variable?

A power series of a function of 2 variables has two independent variables, while a power series of a function of 1 variable has only one independent variable. This means that the terms in a power series of a function of 2 variables will have two variables raised to different powers, while the terms in a power series of a function of 1 variable will have one variable raised to different powers.

3. What are the applications of power series of a function of 2 variables?

Power series of a function of 2 variables have various applications in mathematics, physics, and engineering. They are used to approximate functions, solve differential equations, and describe complex systems such as fluid flow and heat transfer.

4. How do you determine the convergence of a power series of a function of 2 variables?

The convergence of a power series of a function of 2 variables can be determined by using the Cauchy-Hadamard theorem, which states that the radius of convergence of a power series is equal to the reciprocal of the limit superior of the nth root of the coefficients. Additionally, the ratio test and the root test can also be used to determine convergence.

5. Can a power series of a function of 2 variables have a finite radius of convergence?

Yes, a power series of a function of 2 variables can have a finite radius of convergence. This means that the series will only converge for a certain range of values of the two variables. Beyond this range, the series will diverge. This is in contrast to a power series of a function of 1 variable, which can have either a finite or infinite radius of convergence.

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