Prove that log(1 + 𝑒
𝑥) = 𝑙𝑜𝑔2 +
𝑥
2
+
𝑥
2
8
−
𝑥
4
192
+ ⋯ … … … …

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!

Prove that log(1 + e^x) = log 2 + x/2 + x^2/8 − x^4/192 + ⋯

This detailed guide walks through the proof of the logarithmic expansion log(1 + e^x) = log 2 + x/2 + x^2/8 − x^4/192 + ⋯ using the Taylor series. Understand key mathematical concepts such as logarithmic and exponential functions, and how series expansions are derived. Suitable for undergraduate math students.

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