Problem 38 - The Calculus with Analytic Geometry

Question

If $$ y=\frac{1}{1-2 x} $$ prove by mathematical induction that $$ D_{x}^{n} y=\frac{2^{n} n !}{(1-2 x)^{n+1}} $$

Step-by-Step Solution

Answer

The statement is proven true by induction.

Step 1 - Base Case (n=0)

First, check if the statement holds for the base case when n=0. For n=0,\[ D_{x}^{0} y = y = \frac{1}{1-2x} \]We need to confirm that this is equal to\[ \frac{2^{0} 0!}{(1-2x)^{0+1}} = \frac{1}{1-2x} \]Indeed, both expressions are equal, so the base case is true.

Step 2 - Inductive Hypothesis

Assume the statement holds for some positive integer k, i.e.,\[ D_{x}^{k} y = \frac{2^{k} k!}{(1-2x)^{k+1}} \]This is our inductive hypothesis.

Step 3 - Inductive Step

We need to prove that the statement holds for k+1, i.e.,\[ D_{x}^{k+1} y = \frac{2^{k+1} (k+1)!}{(1-2x)^{k+2}} \]Using the inductive hypothesis:\[ D_{x}^{k} y = \frac{2^{k} k!}{(1-2x)^{k+1}} \]By differentiating both sides with respect to x:\[ D_{x}^{k+1} y = D_{x} \left( \frac{2^{k} k!}{(1-2x)^{k+1}} \right) \]

Step 4 - Differentiation

To differentiate \( \frac{2^{k} k!}{(1-2x)^{k+1}} \) with respect to x, use the chain rule:\[ D_{x} \left( \frac{2^{k} k!}{(1-2x)^{k+1}} \right) = 2^{k} k! \cdot D_{x} \left( (1-2x)^{-(k+1)} \right) \]Use the chain rule again on \( (1-2x)^{-(k+1)} \):\[ D_{x} \left( (1-2x)^{-(k+1)} \right) = -(k+1) \cdot (1-2x)^{-(k+2)} \cdot (-2) = 2(k+1) \cdot (1-2x)^{-(k+2)} \]

Step 5 - Simplifying the Result

Combine the results of differentiation:\[ D_{x}^{k+1} y = 2^{k} k! \cdot 2(k+1) \cdot (1-2x)^{-(k+2)} \]Simplify the expression:\[ D_{x}^{k+1} y = 2^{k+1} (k+1)! \cdot (1-2x)^{-(k+2)} = \frac{2^{k+1} (k+1)!}{(1-2x)^{k+2}} \]This shows that if the inductive hypothesis holds for k, it also holds for k+1.

Step 6 - Conclusion

By the principle of mathematical induction, since the base case is true and the inductive step has been proven, the statement is true for all natural numbers n.

Key Concepts

headline of the respective core concept

In calculus, differentiation is a fundamental concept that involves finding the rate at which a function changes at any given point. This process is crucial in many areas of mathematics and science. For the function given in the exercise, we are asked to prove a more complex result involving differentiation using mathematical induction.
Let's break this down.
First, differentiation allows us to find the derivative of a function, which tells us how the function's output value changes as its input value changes. For example, the derivative of the function \( y = \frac{1}{1-2x} \) with respect to \(x\) helps us understand how \(y\) varies when \(x\) changes.
When we differentiate a function multiple times, we call these higher-order derivatives. In the problem, we are using induction to prove a formula for the \(n\)-th derivative of \( y \). This technique is commonly used to handle such recursive definitions and extend results from a single derivative to multiple derivatives.

headline of the respective core concept

The chain rule is a powerful tool in differentiation, especially when dealing with composite functions. A composite function is one in which a function is applied within another function, such as \( (1-2x)^{-(k+1)} \) in our problem.
To differentiate a composite function like this, we need to apply the chain rule. The chain rule states that the derivative of a composite function \( f(g(x)) \) is given by \( f'(g(x)) \times g'(x) \).
Let's see how this applies to our exercise. When differentiating \( (1-2x)^{-(k+1)} \) with respect to \( x \), we view this as an outer function \( (u)^{-(k+1)} \) where \( u = 1-2x \) is the inner function.
Using the chain rule, we first differentiate the outer function with respect to \( u \), which gives \( -(k+1)(u)^{-(k+2)} \).
Next, we differentiate the inner function \( u = 1-2x \) with respect to \( x \), noting that \( D_x(1-2x) = -2 \).
By multiplying these results together, we get \( 2(k+1)(1-2x)^{-(k+2)} \). This step is crucial for proving the inductive step in our exercise.

headline of the respective core concept

Mathematical induction is a method used to prove statements about all natural numbers. It's particularly useful for proving claims about sequences, series, and iterative processes. Our goal is to prove that the given formula for \( D_x^n y \) holds for all natural numbers \( n \).
Induction involves two main steps: the base case and the inductive step.

• Base Case: Verify that the statement is true for the initial value of \( n \), usually \( n = 0 \) or \( n = 1 \). In our exercise, for \( n=0 \), it simplifies to verifying the original function \( y = \frac{1}{1-2x} \).


• Inductive Hypothesis: Assume the statement is true for some positive integer \( k \). This is our assumption to leverage for the next step.


• Inductive Step: Prove that if the statement holds for \( k \), it also holds for \( k+1 \).

In the given exercise, the inductive hypothesis is that the formula \( D_x^k y = \frac{2^k k!}{(1-2x)^{k+1}} \) holds for \( k \). We then show that \( D_x^{k+1} y \) follows the formula as well. This involves differentiating the given form and simplifying it.
Once both the base case and inductive step are verified, we can conclude that the formula is valid for all natural numbers \( n \) by induction.