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In Exercises find \(\partial f / \partial x\) and \(\partial f / \partial y\). $$f(x, y)=\sum_{n=0}^{\infty}(x y)^{n} \quad(|x y| < 1)$$

Short Answer

Expert verified
\( \frac{\partial f}{\partial x} = \frac{y}{(1 - xy)^2} \) and \( \frac{\partial f}{\partial y} = \frac{x}{(1 - xy)^2} \).

Step by step solution

01

Understanding the Function Definition

The function given is a series: \( f(x, y) = \sum_{n=0}^{\infty}(x y)^{n} \). This is a geometric series with the common ratio \( xy \) and the condition \( |xy| < 1 \) ensures convergence.
02

Expressing the Geometric Series

Recognize the series \( \sum_{n=0}^{\infty}(xy)^{n} \) is a geometric series which can be summed up using the formula for an infinite geometric series \( \frac{1}{1 - r} \) where \( r = xy \). Thus, \( f(x, y) = \frac{1}{1 - xy} \) for \( |xy| < 1 \).
03

Finding the Partial Derivative \( \frac{\partial f}{\partial x} \)

To find \( \frac{\partial f}{\partial x} \), differentiate \( f(x, y) = \frac{1}{1 - xy} \) with respect to \( x \) while treating \( y \) as a constant. The derivative of \( \frac{1}{1 - xy} \) using the chain rule is \( \frac{y}{(1 - xy)^2} \), leading to \( \frac{\partial f}{\partial x} = \frac{y}{(1 - xy)^2} \).
04

Finding the Partial Derivative \( \frac{\partial f}{\partial y} \)

To find \( \frac{\partial f}{\partial y} \), differentiate \( f(x, y) = \frac{1}{1 - xy} \) with respect to \( y \) while treating \( x \) as a constant. Again using the chain rule, the derivative is \( \frac{x}{(1 - xy)^2} \), so \( \frac{\partial f}{\partial y} = \frac{x}{(1 - xy)^2} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Geometric Series
A geometric series is a sum of the form \(a + ar + ar^2 + ar^3 + \ldots\). In our exercise, the geometric series is expressed as \( \sum_{n=0}^{\infty}(xy)^{n} \). Here, \(xy\) acts as the common ratio \(r\) for each term in the series. This series is termed 'geometric' because each subsequent term is found by multiplying the previous term by \(r\).

For the series to sum to a finite value, a convergence condition must be satisfied: \(|xy| < 1\). This ensures that the terms get smaller and approach zero as \(n\) increases, allowing the series to sum to a finite value. The sum of an infinite geometric series is calculated using the formula: \( \frac{1}{1-r} \), where \(r = xy\). Thus, the given function can be simplified to \( f(x, y) = \frac{1}{1 - xy} \) under the condition \(|xy| < 1\).

This simplification is crucial as it makes it easier to work with the function, especially when taking derivatives.
Chain Rule
The chain rule is a fundamental technique in calculus used for differentiation. It is particularly useful when dealing with composite functions. A composite function is one where the output of one function becomes the input of another. In simpler terms, if you have a function \(g(z)\) and \(z = h(x)\), the chain rule allows you to differentiate \(f(h(x))\) using \(\frac{df}{dx} = \frac{df}{dz} \cdot \frac{dz}{dx}\).

In the solution, the function \(f(x, y) = \frac{1}{1 - xy}\) is treated as a composite function. When we differentiate with respect to \(x\) and \(y\), we apply the chain rule. The differentiation leads to expressions \( \frac{y}{(1 - xy)^2} \) for \(\frac{\partial f}{\partial x}\) and \( \frac{x}{(1 - xy)^2} \) for \(\frac{\partial f}{\partial y}\).

Here's a simple breakdown:
  • Differentiating \(1-xy\) inside the function leads \(\frac{d}{dx}(1 - xy) = -y\) and \(\frac{d}{dy}(1 - xy) = -x\).
  • Apply this using the chain rule as \(\frac{d}{du} \left( \frac{1}{u} \right) \cdot \frac{du}{dx}\) and \(\frac{du}{dy}\).
This allows us to elegantly find the partial derivatives.
Convergence Condition
In the realm of infinite series, understanding when a series converges is paramount. Convergence within series representations, such as geometric series, is what ensures that they can be equated to finite sums.

The convergence condition \(|xy| < 1\) is key here. It establishes the parameter within which the series converges, i.e., results in a finite sum. This condition restricts the values \(x\) and \(y\) can assume, ensuring that \(xy\) remains in the interval \(-1, 1\).

Why is this important? The central idea is to avoid divergence, where terms grow indefinitely making it impossible to obtain a finite sum. Convergent series maintain a balance, ensuring terms decrease sufficiently quickly to approach an actual sum. In this specific example, the function \(f(x, y)\) can solely be represented and used in its simplified form \(\frac{1}{1 - xy}\) when the convergence criterion is met.

Understanding such conditions is crucial when dealing with series in calculus and its practical applications.

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Most popular questions from this chapter

Use a CAS to perform the following steps: a. Plot the function over the given rectangle. b. Plot some level curves in the rectangle. c. Calculate the function's first partial derivatives and use the CAS equation solver to find the critical points. How do the critical points relate to the level curves plotted in part (b)? Which critical points, if any, appear to give a saddle point? Give reasons for your answer. d. Calculate the function's second partial derivatives and find the discriminant \(f_{x x} f_{y y}-f_{x y}^{2}\) e. Using the max-min tests, classify the critical points found in part (c). Are your findings consistent with your discussion in part (c)? $$\begin{aligned}&f(x, y)=2 x^{4}+y^{4}-2 x^{2}-2 y^{2}+3,-3 / 2 \leq x \leq 3 / 2\\\ &-3 / 2 \leq y \leq 3 / 2\end{aligned}$$

Find the maximum value that \(f(x, y, z)=x^{2}+2 y-z^{2}\) can have on the line of intersection of the planes \(2 x-y=0\) and \(y+z=0.\)

Value of a \(2 \times 2\) determinant If \(|a|\) is much greater than \(|b| .|c| .\) and \(|d| .\) to which of \(a, b, c,\) and \(d\) is the value of the determinant $$f(a, b, c, d)=\left|\begin{array}{ll}a & b \\\c & d \end{array}\right|$$ most sensitive? Give reasons for your answer.

Find the dimensions of the rectangular box of maximum volume that can be inscribed inside the sphere \(x^{2}+y^{2}+z^{2}=4\).

a. Around the point \((1,0),\) is \(f(x, y)=x^{2}(y+1)\) more sensitive to changes in \(x\) or to changes in \(y ?\) Give reasons for your answer. b. What ratio of \(d x\) to \(d y\) will make \(d f\) equal zero at (1,0)\(?\)

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