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Answer :
We want to find the first four nonzero terms of the Maclaurin series of
[tex]$$
f(x) = e^{6x} \sin x.
$$[/tex]
The Maclaurin series expansion of a function is its power series expansion about [tex]\( x=0 \)[/tex].
Step 1. Write the known series for each factor.
The series for the exponential function is
[tex]$$
e^{6x} = 1 + 6x + \frac{(6x)^2}{2!} + \frac{(6x)^3}{3!} + \frac{(6x)^4}{4!} + \cdots
= 1 + 6x + 18 x^2 + 36 x^3 + 54 x^4 + \cdots.
$$[/tex]
The series for the sine function is
[tex]$$
\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots
= x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots.
$$[/tex]
Step 2. Multiply the series for [tex]\(e^{6x}\)[/tex] and [tex]\(\sin x\)[/tex].
Since the series for [tex]\(\sin x\)[/tex] begins with [tex]\(x\)[/tex], the lowest degree term in the product will be of degree 1.
We expand carefully to find the terms up to [tex]\(x^4\)[/tex]:
1. Term in [tex]\(x\)[/tex]:
Multiply the constant in [tex]\(e^{6x}\)[/tex] by the [tex]\(x\)[/tex] term in [tex]\(\sin x\)[/tex]:
[tex]$$
1 \cdot x = x.
$$[/tex]
2. Term in [tex]\(x^2\)[/tex]:
Multiply the [tex]\(6x\)[/tex] term (from [tex]\(e^{6x}\)[/tex]) by the [tex]\(x\)[/tex] term (from [tex]\(\sin x\)[/tex]):
[tex]$$
6x \cdot x = 6x^2.
$$[/tex]
3. Term in [tex]\(x^3\)[/tex]:
There are two contributions:
- The constant [tex]\(1\)[/tex] from [tex]\(e^{6x}\)[/tex] times the [tex]\(-\frac{x^3}{6}\)[/tex] term from [tex]\(\sin x\)[/tex]:
[tex]$$
1 \cdot \left(-\frac{x^3}{6}\right) = -\frac{x^3}{6}.
$$[/tex]
- The [tex]\(18x^2\)[/tex] term from [tex]\(e^{6x}\)[/tex] times the [tex]\(x\)[/tex] term from [tex]\(\sin x\)[/tex]:
[tex]$$
18x^2 \cdot x = 18x^3.
$$[/tex]
Summing the contributions, we obtain:
[tex]$$
-\frac{x^3}{6} + 18x^3 = \left(18 - \frac{1}{6}\right)x^3 = \frac{107}{6} x^3.
$$[/tex]
4. Term in [tex]\(x^4\)[/tex]:
The contributions are:
- The [tex]\(6x\)[/tex] term from [tex]\(e^{6x}\)[/tex] times the [tex]\(-\frac{x^3}{6}\)[/tex] term from [tex]\(\sin x\)[/tex]:
[tex]$$
6x \cdot \left(-\frac{x^3}{6}\right) = -x^4.
$$[/tex]
- The [tex]\(36x^3\)[/tex] term from [tex]\(e^{6x}\)[/tex] times the [tex]\(x\)[/tex] term from [tex]\(\sin x\)[/tex]:
[tex]$$
36x^3 \cdot x = 36x^4.
$$[/tex]
Adding these, we get:
[tex]$$
-x^4 + 36x^4 = 35x^4.
$$[/tex]
Step 3. Write the final answer.
The first four nonzero terms in the Maclaurin series for [tex]\( f(x)=e^{6x} \sin x \)[/tex] are
[tex]$$
x + 6x^2 + \frac{107}{6}x^3 + 35x^4 + \cdots.
$$[/tex]
Comparing with the given options, the correct answer is:
B. [tex]\(x+ 6x^2+ \frac{107}{6}x^3+ 35x^4+\cdots\)[/tex].
[tex]$$
f(x) = e^{6x} \sin x.
$$[/tex]
The Maclaurin series expansion of a function is its power series expansion about [tex]\( x=0 \)[/tex].
Step 1. Write the known series for each factor.
The series for the exponential function is
[tex]$$
e^{6x} = 1 + 6x + \frac{(6x)^2}{2!} + \frac{(6x)^3}{3!} + \frac{(6x)^4}{4!} + \cdots
= 1 + 6x + 18 x^2 + 36 x^3 + 54 x^4 + \cdots.
$$[/tex]
The series for the sine function is
[tex]$$
\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots
= x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots.
$$[/tex]
Step 2. Multiply the series for [tex]\(e^{6x}\)[/tex] and [tex]\(\sin x\)[/tex].
Since the series for [tex]\(\sin x\)[/tex] begins with [tex]\(x\)[/tex], the lowest degree term in the product will be of degree 1.
We expand carefully to find the terms up to [tex]\(x^4\)[/tex]:
1. Term in [tex]\(x\)[/tex]:
Multiply the constant in [tex]\(e^{6x}\)[/tex] by the [tex]\(x\)[/tex] term in [tex]\(\sin x\)[/tex]:
[tex]$$
1 \cdot x = x.
$$[/tex]
2. Term in [tex]\(x^2\)[/tex]:
Multiply the [tex]\(6x\)[/tex] term (from [tex]\(e^{6x}\)[/tex]) by the [tex]\(x\)[/tex] term (from [tex]\(\sin x\)[/tex]):
[tex]$$
6x \cdot x = 6x^2.
$$[/tex]
3. Term in [tex]\(x^3\)[/tex]:
There are two contributions:
- The constant [tex]\(1\)[/tex] from [tex]\(e^{6x}\)[/tex] times the [tex]\(-\frac{x^3}{6}\)[/tex] term from [tex]\(\sin x\)[/tex]:
[tex]$$
1 \cdot \left(-\frac{x^3}{6}\right) = -\frac{x^3}{6}.
$$[/tex]
- The [tex]\(18x^2\)[/tex] term from [tex]\(e^{6x}\)[/tex] times the [tex]\(x\)[/tex] term from [tex]\(\sin x\)[/tex]:
[tex]$$
18x^2 \cdot x = 18x^3.
$$[/tex]
Summing the contributions, we obtain:
[tex]$$
-\frac{x^3}{6} + 18x^3 = \left(18 - \frac{1}{6}\right)x^3 = \frac{107}{6} x^3.
$$[/tex]
4. Term in [tex]\(x^4\)[/tex]:
The contributions are:
- The [tex]\(6x\)[/tex] term from [tex]\(e^{6x}\)[/tex] times the [tex]\(-\frac{x^3}{6}\)[/tex] term from [tex]\(\sin x\)[/tex]:
[tex]$$
6x \cdot \left(-\frac{x^3}{6}\right) = -x^4.
$$[/tex]
- The [tex]\(36x^3\)[/tex] term from [tex]\(e^{6x}\)[/tex] times the [tex]\(x\)[/tex] term from [tex]\(\sin x\)[/tex]:
[tex]$$
36x^3 \cdot x = 36x^4.
$$[/tex]
Adding these, we get:
[tex]$$
-x^4 + 36x^4 = 35x^4.
$$[/tex]
Step 3. Write the final answer.
The first four nonzero terms in the Maclaurin series for [tex]\( f(x)=e^{6x} \sin x \)[/tex] are
[tex]$$
x + 6x^2 + \frac{107}{6}x^3 + 35x^4 + \cdots.
$$[/tex]
Comparing with the given options, the correct answer is:
B. [tex]\(x+ 6x^2+ \frac{107}{6}x^3+ 35x^4+\cdots\)[/tex].
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