Using the Maclaurin formula, show that (1+x)^m = 1+mx+

Using the Maclaurin formula, show that $(1+x)^m = 1+mx+\dfrac{m(m-1)}{2!}x^2+\cdots+\dfrac{m(m-1)\cdots(m-n+1)}{n!}x^n+o(x^n)$.

Step 1: Derivatives of $f(x)=(1+x)^m$

Differentiating repeatedly (and confirming the pattern by induction): \[f^{(k)}(x) = m(m-1)(m-2)\cdots(m-k+1)\,(1+x)^{m-k}. \tag{1}\]

Step 2: Evaluate at $x=0$

\[ f^{(k)}(0) = m(m-1)(m-2)\cdots(m-k+1), \qquad k=0,1,\ldots,n. \tag{2} \]

Step 3: Maclaurin formula with Peano remainder

Substituting (2) into $f(x)=\sum_{k=0}^n\dfrac{f^{(k)}(0)}{k!}x^k+o(x^n)$: \[(1+x)^m = \sum_{k=0}^{n}\frac{m(m-1)\cdots(m-k+1)}{k!}x^k + o(x^n). \tag{3}\]

Writing the terms out through $k=n$ gives exactly the stated expansion.

Step 4: Connection to the binomial coefficient

Defining $\dbinom mk=\dfrac{m(m-1)\cdots(m-k+1)}{k!}$, (3) reads compactly as $(1+x)^m=\sum_{k=0}^n\binom mk x^k+o(x^n)$. When $m\in\mathbb N$ and $n=m$, every term with $k>m$ contains the factor $m-m=0$ and vanishes, so this reduces to the classical finite binomial theorem.

Step 5: Verification

\[m=\tfrac12,\ n=3\text{: symbolic expansion of }(1+x)^{1/2}\text{ through }x^3\text{ matches }1+\tfrac x2-\tfrac{x^2}8+\tfrac{x^3}{16}\text{ exactly}\ \checkmark\]