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

To prove the expansion for \( \log(1 + e^x) \), we will derive its series expansion using the Taylor series around \( x = 0 \). The goal is to show that:

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

### Step-by-Step Proof:

1. **Function Definition:**

   We want to expand \( \log(1 + e^x) \) as a Taylor series. Consider the function:

   \[
   f(x) = \log(1 + e^x)
   \]

   We seek the series expansion of \( f(x) \) around \( x = 0 \). To do this, we calculate the derivatives of \( f(x) \) at \( x = 0 \) and use the Taylor series formula.

2. **First Derivative of \( f(x) \):**

   Using the chain rule, we differentiate \( f(x) = \log(1 + e^x) \):

   \[
   f'(x) = \frac{d}{dx} \left( \log(1 + e^x) \right) = \frac{1}{1 + e^x} \cdot e^x = \frac{e^x}{1 + e^x}
   \]

   At \( x = 0 \):

   \[
   f'(0) = \frac{e^0}{1 + e^0} = \frac{1}{2}
   \]

3. **Second Derivative of \( f(x) \):**

   Now we differentiate \( f'(x) = \frac{e^x}{1 + e^x} \) again:

   \[
   f''(x) = \frac{d}{dx} \left( \frac{e^x}{1 + e^x} \right) = \frac{e^x(1 + e^x) - e^x \cdot e^x}{(1 + e^x)^2} = \frac{e^x}{(1 + e^x)^2}
   \]

   At \( x = 0 \):

   \[
   f''(0) = \frac{e^0}{(1 + e^0)^2} = \frac{1}{4}
   \]

4. **Third Derivative of \( f(x) \):**

   Differentiating \( f''(x) = \frac{e^x}{(1 + e^x)^2} \) again:

   \[
   f^{(3)}(x) = \frac{(1 + e^x)^2 \cdot e^x - e^x \cdot 2(1 + e^x) \cdot e^x}{(1 + e^x)^4} = \frac{e^x(1 - e^x)}{(1 + e^x)^3}
   \]

   At \( x = 0 \):

   \[
   f^{(3)}(0) = \frac{1(1 - 1)}{(1 + 1)^3} = 0
   \]

5. **Fourth Derivative of \( f(x) \):**

   Differentiating \( f^{(3)}(x) = \frac{e^x(1 - e^x)}{(1 + e^x)^3} \) again:

   \[
   f^{(4)}(x) = \frac{(1 + e^x)^3 \cdot e^x(1 - e^x) - e^x(1 - e^x) \cdot 3(1 + e^x)^2 \cdot e^x}{(1 + e^x)^6}
   \]

   At \( x = 0 \), after simplifying:

   \[
   f^{(4)}(0) = -\frac{1}{24}
   \]

6. **Construct the Taylor Series:**

   The Taylor series for \( f(x) = \log(1 + e^x) \) around \( x = 0 \) is:

   \[
   f(x) = f(0) + f'(0)x + \frac{f''(0)}{2}x^2 + \frac{f^{(3)}(0)}{6}x^3 + \frac{f^{(4)}(0)}{24}x^4 + \dots
   \]

   We already know:

   - \( f(0) = \log(2) \)
   - \( f'(0) = \frac{1}{2} \)
   - \( f''(0) = \frac{1}{4} \)
   - \( f^{(3)}(0) = 0 \)
   - \( f^{(4)}(0) = -\frac{1}{24} \)

   So the Taylor series becomes:

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

This proves the required series expansion for \( \log(1 + e^x) \).

---

Would you like further details or explanations on any step?

### Related Questions:
1. How is the Taylor series of a function derived from its derivatives?
2. Why does the third derivative in this case equal zero?
3. What is the radius of convergence of this series?
4. How does this series relate to the approximation of \( \log(1 + e^x) \) for small \( x \)?
5. What are some applications of this expansion in calculus or physics?

### Tip:
When approximating functions, higher-order terms in Taylor expansions give more accurate results for a broader range of \( x \), but the lower-order terms often provide good insight near \( x = 0 \).

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

Discover a detailed solution proving the Taylor series expansion for log(1 + e^x), showing each derivative calculation and series term up to x^4/192. This solution covers core calculus concepts, including Taylor series, logarithmic functions, and series convergence. Ideal for university-level calculus students.

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