Question

Let $$ f(x, y)=e^{x+y} $$ By writing $$ f(x, y)=\sum_{r=0}^{\infty} \frac{(x+y)^{r}}{r !} $$ and expanding \((x+y)^{r}\) by means of the binomial theorem, verify that $$ d_{(0,0)}^{(r)} f=\sum_{j=0}^{r}\left(\begin{array}{l} r \\ j \end{array}\right) \frac{\partial^{r} f(0,0)}{\partial x^{j} \partial y^{r-j}}(d x)^{j}(d y)^{r-j} $$

Short answer

Based on the given information, our goal is to verify that the equation: $$ d_{(0,0)}^{(r)} f = \sum_{j=0}^{r} \binom{r}{j} \frac{\partial^r f(0,0)}{\partial x^j \partial y^{r-j}} (dx)^j (dy)^{r-j}. $$ is true for \(f(x, y) = e^{x+y}\). To verify the equation, we first wrote the function \(f(x, y)\) as a power series using the standard power series expansion for the exponential function. We then expanded \((x+y)^r\) using the binomial theorem and incorporated it into the power series. After that, we computed the mixed partial derivatives of the function \(f(x, y)\) at \((0, 0)\) and found the derivatives to be equal to 1. Finally, by comparing the coefficients of the given equation with the power series and binomial expansion of \(f(x, y)\), we were able to confirm that the given equation holds true for \(f(x, y) = e^{x+y}\).

Step-by-step solution

Write the function as a power series

Given the function \(f(x, y) = e^{x+y}\), we can write it as a power series using the standard power series expansion for the exponential function: $$ f(x, y) = \sum_{r=0}^{\infty} \frac{(x+y)^r}{r!}. $$

Expand \((x+y)^r\) using the binomial theorem

Using the binomial theorem, we can expand \((x+y)^r\) as follows: $$ (x+y)^r = \sum_{j=0}^{r} \binom{r}{j} x^j y^{r-j}, $$ where \(\binom{r}{j} = \frac{r!}{j!(r-j)!}\) is the binomial coefficient.

Incorporate the binomial expansion into the power series

By combining the power series expansion and the binomial expansion, we get: $$ f(x, y) = \sum_{r=0}^{\infty} \frac{1}{r!} \sum_{j=0}^{r} \binom{r}{j} x^j y^{r-j}. $$

Compute the mixed partial derivatives of \(f(x, y)\) at \((0, 0)\)

To compute the mixed partial derivatives of \(f(x, y)\) at \((0, 0)\), we first notice that the partial derivatives of the exponential function are also exponential functions. Thus, for any non-negative integers \(p, q\), we have: $$ \frac{\partial^{p+q} f(x,y)}{\partial x^p \partial y^q} = e^{x+y}. $$ Applying these mixed partial derivatives at \((0, 0)\), we get: $$ \frac{\partial^{p+q} f(0,0)}{\partial x^p \partial y^q} = e^{0+0} = 1. $$

Verify the given equation

The given equation to be verified is: $$ d_{(0,0)}^{(r)} f = \sum_{j=0}^{r} \binom{r}{j} \frac{\partial^r f(0,0)}{\partial x^j \partial y^{r-j}} (dx)^j (dy)^{r-j}. $$ From the previous step, we know that \(\frac{\partial^r f(0,0)}{\partial x^j \partial y^{r-j}} = 1\) for all \(j\). Therefore, the given equation becomes: $$ d_{(0,0)}^{(r)} f = \sum_{j=0}^{r} \binom{r}{j} (dx)^j (dy)^{r-j}. $$ Comparing this with the power series and binomial expansion of \(f(x, y)\) (from Steps 1 and 2), we can see that the coefficients of each power of \(x\) and \(y\) match in both expressions. Thus, we have verified that the given equation holds true.

Key concepts

Power Series

In the realm of real analysis, a power series is an infinite series of the form \[\sum_{n=0}^{\infty} a_n(x-c)^n\], where \(a_n\) represents the coefficient of the nth term, \(x\) is the variable, and \(c\) is the center of the series. A power series converges within its radius of convergence, which means the series sums up to a well-defined value within a certain interval around the center \(c\).

For example, the exponential function \(e^x\) can be expressed as a power series centered at \(c=0\), known as its Maclaurin series: \[e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}\]. Here, the coefficients \(a_n\) are \(1/n!\). Understanding power series is crucial in analyzing functions and their behavior at or near a specific point.

Binomial Theorem

The binomial theorem is a foundational formula for expanding powers of a binomial expression like \( (x+y)^n \), where \(n\) is a non-negative integer. According to this theorem, the binomial expression can be expanded as \[ (x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k \], where the binomial coefficient \(\binom{n}{k}\) - also written as \(C(n, k)\) or \(nCk\) - equals \(\frac{n!}{k!(n-k)!}\).

This theorem not only provides a way to algebraically expand binomials but also is used in probability theory, numerical analysis, and in this context, in the study of power series and their derivatives. It offers insight into the coefficients of each term in a multivariable power series.

Partial Derivatives

In multivariable calculus, partial derivatives play a pivotal role. They represent the derivative of a function with respect to one variable while holding the other variables constant. Symbolically, for a function \( f(x, y) \), the partial derivative with respect to \( x \) at a point \( (a, b) \) is denoted as \( \frac{\partial f}{\partial x}(a, b) \).

Partial derivatives are the foundation for gradient vectors, Jacobian matrices, and higher-order derivatives like the mixed partial derivatives used in the Taylor series for multivariable functions. In the context of our exercise, evaluating these derivatives at a point provides the necessary coefficients for the power series expansion of the function.

Exponential Function

The exponential function, denoted as \( e^x \), is one of the most important functions in mathematics, characterized by the fact that its derivative is proportional to the function itself. Its Maclaurin series expansion, \[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \], is fundamental in calculus, differential equations, and real analysis. This function grows rapidly, which is captured by the exponential growth in phenomena such as population growth, radioactive decay, and continuous compounding in finance.

The exponential function's unique property of being its own derivative means that all partial derivatives of \( e^{x+y} \) evaluated at any point will equal the function's value at that point, which simplifies many calculations in higher-level calculus and analysis.