AHL 5.19—Maclaurin series

The Maclaurin series is defined as $f(x)={\sum }_{n=0}^{\infty }{a}_n{x}^n$ where the coefficients $a_n$ are determined by matching derivatives at $x=0$.

Differentiating $P(x)=a_0+a_{1}x+a_2{x}^2+\cdots$ exactly $n$ times and setting $x=0$ gives $P^{(n)}(0)=n!\cdot a_n$. Therefore $a_n=\frac{f^{(n)}(0)}{n!}$.

The $n$th derivative of $x^n$ is $n!$ (a constant), so dividing by $n!$ correctly isolates the coefficient $a_n$. Without dividing by $n!$, the polynomial would overshoot every derivative value by a factor of $n!$.

The coefficient of $x^n$ is $\frac{f^{(n)}(0)}{n!}$, confirming the second option.

We need to identify the correct formula for the $n$th coefficient in a Maclaurin series. This comes directly from the definition $f(x)={\sum }_{n=0}^{\infty }\frac{f^{(n)}(0)}{n!}{x}^n$.

The option $f^{(n)}(0)$ omits the crucial $n!$ denominator. If the coefficient were simply $f^{(n)}(0)$, then matching $P^{(n)}(0)$ to $f^{(n)}(0)$ would be off by a factor of $n!$ due to the repeated differentiation of $x^n$. This is incorrect.

The option $\frac{f^{(n)}(0)}{(n-1)!}$ uses $(n-1)!$ instead of $n!$. The correct divisor is $n!$ because the $n$th derivative of $x^n$ is $n!$, not $(n-1)!$. This option is off by a factor of $n$.

Multiplying by $n!$ instead of dividing by it inverts the relationship completely. This would make coefficients grow extremely rapidly and the series would diverge for even small $x$.

The only correct formula is $\frac{f^{(n)}(0)}{n!}$, which properly accounts for the $n!$ arising from repeated differentiation.

Starting from the standard series $e^u=1+u+\frac{u^2}{2!}+\frac{u^3}{3!}+\cdots$, substitute $u=2x$ to get the series for $e^{2x}$.

The $x^3$ term in $e^{2x}$ comes from the $u^3$ term: $\frac{(2x{)}^3}{3!}=\frac{8x^3}{6}=\frac{4x^3}{3}$

The coefficient of $x^3$ is $\frac{4}{3}$.

The coefficient of $x^3$ in $e^{2x}$ is $\frac{f^{\prime \prime \prime }(0)}{3!}$, where $f(x)=e^{2x}$. Alternatively, the $x^3$ term is $\frac{(2x{)}^3}{3!}$.

This would correspond to $e^x$ evaluated at the $x^3$ term: $\frac{1}{3!}=\frac{1}{6}$, not $\frac{1}{3}$. Either way, it doesn't account for the factor of $2^3=8$ from the substitution $u=2x$.

This could arise from writing $\frac{(2x{)}^3}{3}=\frac{8x^3}{3}$, incorrectly using $3$ instead of $3!=6$ in the denominator. The correct denominator is $3!=6$, giving $\frac{8}{6}=\frac{4}{3}$.

This might arise from taking $\frac{2}{3!}\cdot x^3$, using $2^1$ rather than $2^3=8$ in the numerator. The substitution $u=2x$ means the $u^3$ term contributes $(2x{)}^3=8x^3$.

The correct coefficient is $\frac{8}{6}=\frac{4}{3}$, from $\frac{(2x{)}^3}{3!}=\frac{8x^3}{6}$.