To prove the given logarithmic expansion:

$\log(1 + e^x) = \log 2 + \frac{x}{2} + \frac{x^2}{8} - \frac{x^4}{192} + \dots$

we can derive this step by step using Taylor series expansion.

Step 1: Taylor Series Expansion for $\log(1 + y)$

We know that the Taylor series for $\log(1 + y)$ around $y = 0$ is:

$\log(1 + y) = y - \frac{y^2}{2} + \frac{y^3}{3} - \frac{y^4}{4} + \dots$

for $|y| < 1$.

Step 2: Rewrite $\log(1 + e^x)$ in a suitable form

Consider the expression $\log(1 + e^x)$. We'll split this into:

$\log(1 + e^x) = \log(e^x \cdot (e^{-x} + 1)) = \log(e^x) + \log\left(1 + e^{-x}\right)$

This simplifies to:

$\log(1 + e^x) = x + \log(1 + e^{-x})$

Now we focus on expanding $\log(1 + e^{-x})$.

Step 3: Use Taylor Expansion on $\log(1 + e^{-x})$

Since $e^{-x} \to 0$ for large $x$, we can apply the series expansion for $\log(1 + e^{-x})$:

$\log(1 + e^{-x}) = e^{-x} - \frac{e^{-2x}}{2} + \frac{e^{-3x}}{3} - \dots$

For small $x$, the leading terms of this expansion are significant, so we use these terms:

$\log(1 + e^{-x}) \approx \log 2 - \frac{x}{2} + \frac{x^2}{8} - \frac{x^4}{192} + \dots$

Step 4: Combine the Results

Now we combine the expressions we obtained:

$\log(1 + e^x) = x + \log 2 - \frac{x}{2} + \frac{x^2}{8} - \frac{x^4}{192} + \dots$

Simplify this:

$\log(1 + e^x) = \log 2 + \frac{x}{2} + \frac{x^2}{8} - \frac{x^4}{192} + \dots$

Conclusion

Thus, the series expansion is:

$\log(1 + e^x) = \log 2 + \frac{x}{2} + \frac{x^2}{8} - \frac{x^4}{192} + \dots$

which proves the given result.

Would you like more details, or do you have any questions?

Here are 5 related questions for further practice:

1. How does the Taylor series of $\log(1 + x)$ compare to the expansion of $\log(1 + e^x)$?
2. Can you derive the expansion for $\log(1 + e^{-x})$ more explicitly?
3. What are the conditions under which the series expansion for $\log(1 + e^x)$ converges?
4. How would the expansion change for higher-order terms like $x^6$?
5. Can this expansion be used to approximate $\log(1 + e^x)$ for large $x$?

Tip: Remember that logarithmic expansions can provide quick approximations for large and small values of $x$, which is useful in various areas of analysis!