(a) Given $$ f(x)=e^{\iota x} $$, obtain the Maclaurin expansion for $$ f(x) $$ up to, and including, the term in $$ x^{3} $$. (b) On a suitable domain, let $$ g(x)= an x $$. (i) Show that the third derivative of $$ g(x) $$ is given by $$ g^{\prime \prime \prime}(x)=2 \sec ^{4} x+4 an ^{2} x \sec ^{2} x $$. (ii) Hence obtain the Maclaurin expansion for $$ g(x) $$ up to and including the term in $$ x^{3} $$. (c) Hence, or otherwise, obtain the Maclaurin expansion for $$ e^{2 x} an x $$ up to, and including, the term in $$ x^{3} $$. (d) Write down the first three non-zero terms in the Maclaurin expansion for $$ 2 e^{2 x} an x+e^{2 x} \sec ^{2} x $$.

Question

(a) Given $$ f(x)=e^{\iota x} $$, obtain the Maclaurin expansion for $$ f(x) $$ up to, and including, the term in $$ x^{3} $$. (b) On a suitable domain, let $$ g(x)=	an x $$. (i) Show that the third derivative of $$ g(x) $$ is given by $$ g^{\prime \prime \prime}(x)=2 \sec ^{4} x+4 	an ^{2} x \sec ^{2} x $$. (ii) Hence obtain the Maclaurin expansion for $$ g(x) $$ up to and including the term in $$ x^{3} $$. (c) Hence, or otherwise, obtain the Maclaurin expansion for $$ e^{2 x} 	an x $$ up to, and including, the term in $$ x^{3} $$. (d) Write down the first three non-zero terms in the Maclaurin expansion for $$ 2 e^{2 x} 	an x+e^{2 x} \sec ^{2} x $$.

UpStudy AI · Accepted Answer

### Problem (a)

Given $$ f(x) = e^{\iota x} $$, we want to find the Maclaurin expansion up to the $$ x^3 $$ term.

The Maclaurin series for $$ e^{\iota x} $$ is given by:
$$
e^{\iota x} = 1 + \iota x + \frac{(\iota x)^2}{2!} + \frac{(\iota x)^3}{3!} + \cdots
$$

Calculating each term up to $$ x^3 $$:

1. **Constant term:**
   $$
   1
   $$

2. **Linear term:**
   $$
   \iota x
   $$

3. **Quadratic term:**
   $$
   \frac{(\iota x)^2}{2!} = \frac{\iota^2 x^2}{2} = \frac{(-1) x^2}{2} = -\frac{x^2}{2}
   $$
   (since $$ \iota^2 = -1 $$)

4. **Cubic term:**
   $$
   \frac{(\iota x)^3}{3!} = \frac{\iota^3 x^3}{6} = \frac{(-\iota) x^3}{6} = -\frac{\iota x^3}{6}
   $$
   (since $$ \iota^3 = -\iota $$)

Putting it all together, the Maclaurin expansion up to $$ x^3 $$ is:
$$
f(x) = 1 + \iota x - \frac{x^2}{2} - \frac{\iota x^3}{6}
$$

**Final Answer:**

Problem a Answer:

Expanding term by term, up to and including $$x^{3}$$, we have

$$
e^{\iota x} = 1 + \iota\,x - \tfrac{1}{2}x^{2} - \tfrac{1}{6}\iota\,x^{3}.
$$
The Maclaurin expansion for $$ f(x) = e^{\iota x} $$ up to the $$ x^3 $$ term is: $$ 1 + \iota x - \frac{x^2}{2} - \frac{\iota x^3}{6} $$

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