Problem 278
Easy Uniform Convergence

Write the first four non-zero terms of the power series

Write the first four non-zero terms of the power series in $x$ for $(1+x)^x$.

Explanation of Method

Rewriting $(1+x)^x=e^{x\ln(1+x)}$ reduces the problem to composing the exponential series with $u=x\ln(1+x)$, which itself starts at $x^2$ (since $\ln(1+x)$ starts at $x^1$); collecting powers of $u$ then gives the coefficients.

  1. Rewrite $(1+x)^x=e^{x\ln(1+x)}$.
  2. Expand $u=x\ln(1+x)=x^2-\frac{x^3}2+\frac{x^4}3-\cdots$.
  3. Expand $e^u=1+u+\frac{u^2}2+\frac{u^3}6+\cdots$ and collect terms.

Step 1 · Rewrite and expand $u$

$$ (1+x)^x = e^{x\ln(1+x)}, \qquad u=x\ln(1+x)=x^2-\frac{x^3}2+\frac{x^4}3-\frac{x^5}4+\cdots $$

Step 2 · Powers of $u$

$$ u^2 = x^4-x^5+\frac{11x^6}{12}+\cdots, \qquad u^3 = x^6+\cdots $$

Step 3 · Collect $e^u=1+u+\frac{u^2}2+\frac{u^3}6+\cdots$

Coefficient of $x^2$: $1$. Coefficient of $x^3$: $-\frac12$. Coefficient of $x^4$: $\frac13+\frac12=\frac56$.


Summary Table

PropertyValue
First four non-zero terms$1,\ x^2,\ -\frac{x^3}2,\ \frac{5x^4}6$
At $x=0$$(1+0)^0=1$
Region of validity$|x|<1$

Additional Notes

Result

$$ (1+x)^x = 1+x^2-\frac{x^3}2+\frac{5x^4}6+\cdots, \quad |x|<1. $$