2.8 Q-3

Question Statement

Prove that:

$$2^{x+h} = 2^x { 1 + (\ln 2)h + \frac{(\ln 2)^2}{2!}h^2 + \frac{(\ln 2)^3}{3!}h^3 + \ldots }$$

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Background and Explanation

This problem uses the Taylor series expansion, which is a method for expressing a function in terms of its derivatives at a specific point. The Taylor series for a function $f(x)$ about a point $x$ is given by:

$$f(x+h) = f(x) + \frac{f'(x)}{1!}h + \frac{f''(x)}{2!}h^2 + \frac{f'''(x)}{3!}h^3 + \ldots$$

In this problem, $f(x) = 2^x$. We’ll compute its derivatives and substitute them into the Taylor series formula.

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Solution

1. Let the function be defined as:
   $f(x) = 2^x.$

2. Differentiate the function repeatedly with respect to $x$:

   • The first derivative:
     $f'(x) = 2^x \ln 2$
   • The second derivative:
     $f''(x) = 2^x (\ln 2)^2$
   • The third derivative:
     $f'''(x) = 2^x (\ln 2)^3$
   • The fourth derivative:
     $f^{(4)}(x) = 2^x (\ln 2)^4$

3. Substitute these derivatives into the Taylor series formula:

   The general Taylor series is:

$$f(x+h) = f(x) + \frac{f'(x)}{1!}h + \frac{f''(x)}{2!}h^2 + \frac{f'''(x)}{3!}h^3 + \ldots$$

For $f(x) = 2^x$, this becomes:

$$2^{x+h} = 2^x + \frac{2^x (\ln 2)}{1!}h + \frac{2^x (\ln 2)^2}{2!}h^2 + \frac{2^x (\ln 2)^3}{3!}h^3 + \ldots$$

4. Factor out $2^x$ to simplify:

$$2^{x+h} = 2^x \left[ 1 + (\ln 2)h + \frac{(\ln 2)^2}{2!}h^2 + \frac{(\ln 2)^3}{3!}h^3 + \ldots \right]$$

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Key Formulas or Methods Used

• Taylor Series:
  $f(x+h) = f(x) + \frac{f'(x)}{1!}h + \frac{f''(x)}{2!}h^2 + \frac{f'''(x)}{3!}h^3 + \ldots$

• Derivative Rules for Exponential Functions:
  For $f(x) = a^x$, the derivative is $f'(x) = a^x \ln a$.

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Summary of Steps

1. Start with $f(x) = 2^x$.
2. Compute the derivatives:
   • $f'(x) = 2^x \ln 2$
   • $f''(x) = 2^x (\ln 2)^2$
   • $f'''(x) = 2^x (\ln 2)^3$, etc.
3. Substitute the derivatives into the Taylor series expansion.
4. Simplify the expression and factor out $2^x$.

Final result:

$$2^{x+h} = 2^x \left[ 1 + (\ln 2)h + \frac{(\ln 2)^2}{2!}h^2 + \frac{(\ln 2)^3}{3!}h^3 + \ldots \right]$$