(a) Find and simplify the Maclaurin expansion, up to and including the term in ,
for:
(i)
(ii) .
(b) Hence find the Maclaurin expansion for up to and including the term in
.

Ask by Perkins Norris. in the United Kingdom

Jan 08,2025

Question

(a) Find and simplify the Maclaurin expansion, up to and including the term in ,
for:
(i)
(ii) .
(b) Hence find the Maclaurin expansion for up to and including the term in
.

Ask by Perkins Norris. in the United Kingdom

Jan 08,2025

UpStudy AI · Accepted Answer

### Problem (a)

**(i) Maclaurin Expansion for $$ e^{2x} $$ up to $$ x^3 $$**

The Maclaurin series for $$ e^z $$ is given by:
$$
e^z = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots
$$
Substituting $$ z = 2x $$:
$$
e^{2x} = 1 + 2x + \frac{(2x)^2}{2!} + \frac{(2x)^3}{3!} + \cdots
$$
Simplifying each term up to $$ x^3 $$:
$$
e^{2x} = 1 + 2x + \frac{4x^2}{2} + \frac{8x^3}{6} + \cdots = 1 + 2x + 2x^2 + \frac{4}{3}x^3
$$

**Final Answer:**
$$
e^{2x} = 1 + 2x + 2x^2 + \frac{4}{3}x^3 + \mathcal{O}(x^4)
$$

---

**(ii) Maclaurin Expansion for $$ \sin 3x $$ up to $$ x^3 $$**

The Maclaurin series for $$ \sin z $$ is:
$$
\sin z = z - \frac{z^3}{3!} + \cdots
$$
Substituting $$ z = 3x $$:
$$
\sin 3x = 3x - \frac{(3x)^3}{6} + \cdots = 3x - \frac{27x^3}{6} = 3x - \frac{9}{2}x^3
$$

**Final Answer:**
$$
\sin 3x = 3x - \frac{9}{2}x^3 + \mathcal{O}(x^5)
$$

---

### Problem (b)

**Maclaurin Expansion for $$ e^{2 \sin 3x} $$ up to $$ x^3 $$**

From part (a)(ii), we have:
$$
\sin 3x = 3x - \frac{9}{2}x^3 + \mathcal{O}(x^5)
$$
Thus,
$$
2 \sin 3x = 6x - 9x^3 + \mathcal{O}(x^5)
$$
Let $$ z = 6x - 9x^3 $$. The Maclaurin series for $$ e^z $$ up to $$ z^3 $$ is:
$$
e^z = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots
$$
Substituting $$ z $$:
$$
e^{2 \sin 3x} = 1 + (6x - 9x^3) + \frac{(6x - 9x^3)^2}{2} + \frac{(6x - 9x^3)^3}{6} + \cdots
$$
Calculating each term up to $$ x^3 $$:

1. **Linear Term:**
   $$
   z = 6x - 9x^3
   $$
2. **Quadratic Term:**
   $$
   \frac{z^2}{2} = \frac{(6x)^2}{2} + \text{higher order terms} = \frac{36x^2}{2} = 18x^2
   $$
3. **Cubic Term:**
   $$
   \frac{z^3}{6} = \frac{(6x)^3}{6} + \text{higher order terms} = \frac{216x^3}{6} = 36x^3
   $$

Adding all contributions:
$$
e^{2 \sin 3x} = 1 + (6x - 9x^3) + 18x^2 + 36x^3
$$
Combine like terms:
$$
e^{2 \sin 3x} = 1 + 6x + 18x^2 + ( -9x^3 + 36x^3 ) = 1 + 6x + 18x^2 + 27x^3 + \mathcal{O}(x^4)
$$

**Final Answer:**
$$
e^{2 \sin 3x} = 1 + 6x + 18x^2 + 27x^3 + \mathcal{O}(x^4)
$$
The Maclaurin expansion for $$ e^{2 \sin 3x} $$ up to $$ x^3 $$ is: $$ 1 + 6x + 18x^2 + 27x^3 $$

(a) Find and simplify the Maclaurin expansion, up to and including the term in ,
for:
(i)
(ii) .
(b) Hence find the Maclaurin expansion for up to and including the term in
.

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(a) Find and simplify the Maclaurin expansion, up to and including the term in , for: (i) (ii) . (b) Hence find the Maclaurin expansion for up to and including the term in .

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