Write the first several terms of the Maclaurin polynomial

Write the first several terms of the Maclaurin polynomial for $y=a^x$, $a>0$.

Step 1: Derivatives at $x=0$

Since $a^x=e^{x\ln a}$, the chain rule gives $\dfrac{d^k}{dx^k}a^x=(\ln a)^k a^x$, so

\[ f^{(k)}(0)=(\ln a)^k. \tag{1} \]

Explicitly: $f(0)=1$, $f'(0)=\ln a$, $f''(0)=(\ln a)^2$, $f'''(0)=(\ln a)^3$, $f^{(4)}(0)=(\ln a)^4$.

Step 2: Maclaurin formula

Substituting into $P_n(x)=\sum_{k=0}^n\dfrac{f^{(k)}(0)}{k!}x^k$: \[a^x = 1+\frac{\ln a}{1!}x+\frac{(\ln a)^2}{2!}x^2+\frac{(\ln a)^3}{3!}x^3+\frac{(\ln a)^4}{4!}x^4+\cdots+\frac{(\ln a)^n}{n!}x^n+R_n(x). \tag{2}\]

Step 3: Lagrange remainder and convergence

\[ R_n(x)=\frac{(\ln a)^{n+1}a^{\xi}}{(n+1)!}\,x^{n+1}, \qquad \xi\text{ strictly between }0\text{ and }x. \tag{3} \]

For any fixed $x$, $a^\xi$ stays bounded while $(n+1)!$ grows without bound, so $R_n(x)\to0$ as $n\to\infty$ for every real $x$; the series converges everywhere, consistent with $a^x=e^{x\ln a}=\sum_{k=0}^\infty\dfrac{(x\ln a)^k}{k!}$.

Step 4: Verification

\[a=2,\ x=0.3,\ n=4\text{: }P_4(0.3)\approx1.23107,\ 2^{0.3}\approx1.23114,\text{ diff}\approx7\times10^{-5}\ \checkmark\]