Detailed Answer

a) We differentiate \( g(x) = e^{3x} \):

\[ g(x) = e^{3x} \quad \Rightarrow \quad g(0) = 1 \]

\[ g'(x) = 3e^{3x} \quad \Rightarrow \quad g'(0) = 3 \]

\[ g''(x) = 9e^{3x} \quad \Rightarrow \quad g''(0) = 9 \]

\[ g'''(x) = 27e^{3x} \quad \Rightarrow \quad g'''(0) = 27 \]

\[ g''''(x) = 81e^{3x} \quad \Rightarrow \quad g''''(0) = 81 \]

Thus, the Maclaurin series up to \(x^4\) is:

\[ g(x) = 1 + 3x + \frac{9}{2}x^2 + \frac{27}{6}x^3 + \frac{81}{24}x^4 + \dots \]

\[ g(x) = 1 + 3x + \frac{9}{2}x^2 + \frac{9}{2}x^3 + \frac{27}{8}x^4 + \dots \]

b) Substituting \(x \to 3x\) in the Maclaurin series for \(e^x\), we get:

\[ e^{3x} = 1 + 3x + \frac{(3x)^2}{2!} + \frac{(3x)^3}{3!} + \frac{(3x)^4}{4!} + \dots \]

\[ e^{3x} = 1 + 3x + \frac{9}{2}x^2 + \frac{9}{2}x^3 + \frac{27}{8}x^4 + \dots \]

Detailed Answer

a) We start by expanding \(e^x\) and \(\cos x\) as Maclaurin series:

\[ e^x = 1 + x + \frac{x^2}{2!} \]

\[ \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} \]

Multiplying these series together:

\[ e^x \cos x = \left(1 + x + \frac{x^2}{2!}\right)\left(1 - \frac{x^2}{2!} + \frac{x^4}{4!}\right) \]

\[ = 1 + x + \frac{x^2}{2} - \frac{x^2}{2} + \dots = 1 + x - \frac{x^3}{3!} \]

b) Now, we compute \( e^x \cos x - 1 - x \):

\[ e^x \cos x - 1 - x = \left(1 + x - \frac{x^3}{3!}\right) - 1 - x \]

\[ = -\frac{x^3}{3!} \]

Thus:

\[ \lim_{x \to 0} \frac{e^x \cos x - 1 - x}{x^2} = \lim_{x \to 0} \frac{-\frac{x^3}{3}}{x^2} = 0. \]

(All the larger powers from here will also disappear as they have a larger power than \(x^2\))

Detailed Answer

To find the first non-zero term in the Maclaurin series of \( g(x) = e^{-x} \cos x - 1 + x - \frac{x^2}{2} \), we need to compute the derivatives of \( g \) at \( x = 0 \) until we find the first non-zero derivative.

\[ g(0) = e^{0} \cos 0 - 1 + 0 - \frac{0^2}{2} = 1 \cdot 1 - 1 + 0 - 0 = 0 \]

\[ g'(x) = \frac{d}{dx} \left( e^{-x} \cos x \right) + \frac{d}{dx} (-1) + \frac{d}{dx} (x) - \frac{d}{dx} \left( \frac{x^2}{2} \right) \]

Using the product rule for \( \frac{d}{dx} \left( e^{-x} \cos x \right) \):

\[ g'(x) = -e^{-x} \cos x - e^{-x} \sin x + 0 + 1 - x \]

Simplifying:

\[ g'(x) = -e^{-x} (\cos x + \sin x) + 1 - x \]

Now, evaluate \( g'(0) \):

\[ g'(0) = -e^{0} (\cos 0 + \sin 0) + 1 - 0 = -1 (1 + 0) + 1 - 0 = -1 + 1 = 0 \]

\[ g''(x) = \frac{d}{dx} \left( -e^{-x} (\cos x + \sin x) \right) + \frac{d}{dx} (1) - \frac{d}{dx} (x) \]

Using the product rule again:

\[ g''(x) = e^{-x} (\cos x + \sin x) - e^{-x} (-\sin x + \cos x) + 0 - 1 \]

Simplifying:

\[ g''(x) = e^{-x} (\cos x + \sin x + \sin x - \cos x) - 1 = e^{-x} (2 \sin x) - 1 \]

Now, evaluate \( g''(0) \):

\[ g''(0) = e^{0} (2 \sin 0) - 1 = 1 \cdot 0 - 1 = -1 \]

Since \( g''(0) = -1 \neq 0 \), the first non-zero term in the Maclaurin series is the \( x^2 \) term.

Substituting the values:

\[ g(x) = 0 + 0 \cdot x + \frac{-1}{2} x^2 + \cdots = -\frac{1}{2} x^2 + \cdots \]

Detailed Answer

a) We start by using the Maclaurin series expansion for \( \sin x \):

\[ \sin x = x - \frac{x^3}{3!} + \dots \]

Using the known expansion \( \ln(1+y) = y - \frac{y^2}{2} + \dots \) for small \( y \), we substitute \( y = \sin x \):

\[ \ln(1 + \sin x) = x - \frac{x^2}{2} + \dots \]

b) To approximate \( \ln 1.5 \), we substitute \( x = \frac{\pi}{6} \) (since \( \sin{\frac{\pi}{6}} = 0.5 \)):

\[ \ln(1 + \sin(\pi/6)) \approx \frac{\pi}{6} - \frac{(\pi/6)^2}{2} \]

Computing values:

\[ \frac{\pi}{6} \approx 0.5236, \quad \left(\frac{\pi}{6}\right)^2 \approx 0.2742 \]

Thus,

\[ \ln 1.5 \approx 0.5236 - \frac{0.2742}{2} = 0.5236 - 0.1371 = 0.3865 \]

Detailed Answer

a) We start by writing the function:

\[ g(x) = \ln(1 + \cos x) \]

Using the chain rule:

\[ g'(x) = \frac{d}{dx} \ln(1+\cos x) = \frac{1}{1+\cos x} \cdot \frac{d}{dx} (\cos x) \]

Since \( \frac{d}{dx} \cos x = -\sin x \), we get:

\[ g'(x) = \frac{-\sin x}{1+\cos x} \]

b) Using the quotient rule where \( u = -\sin x \) and \( v = 1+\cos x \):

\[ g''(x) = \frac{v u' - u v'}{v^2} \]

where:

\[ u' = -\cos x, \quad v' = -\sin x \]

\[ g''(x) = \frac{(1+\cos x)(-\cos x) - (-\sin x)(-\sin x)}{(1+\cos x)^2} \]

\[ = \frac{-\cos x - \cos^2 x - \sin^2 x}{(1+\cos x)^2} \]

Using \( \sin^2 x + \cos^2 x = 1 \):

\[ g''(x) = \frac{-1 - \cos x}{(1+\cos x)^2} \]

Finding the third derivative:

\[ g'''(x) = \frac{dx}{dy}\left(-\frac{1}{1\cos{x}} \right) = \left(-\frac{\sin{(x)}}{(1+\cos{(x)})^2}\right) \]

c) Since \( \cos x \) has a Maclaurin series:

\[ \cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} \]

We substitute into \( g(x) \):

\[ 1+\cos x = 2 - \frac{x^2}{2} + \frac{x^4}{24} \]

Using the logarithm expansion:

\[ \ln(1+y) = y - \frac{y^2}{2} + \frac{y^3}{3} - \frac{y^4}{4} \]

for \( y = \cos x \), we get:

\[ g(x) = \ln \left( 2 - \frac{x^2}{2} + \frac{x^4}{24} \right) \]

Approximating for small \( x \):

\[ g(x) = \ln 2 + \ln \left( 1 + \frac{-x^2/2 + x^4/24}{2} \right) \]

\[ = \ln 2 + \left(\frac{-x^2}{4} + \frac{x^4}{48}\right) - \frac{1}{2} \left(\frac{-x^2}{4} + \frac{x^4}{48}\right)^2 \]

\[ = \ln 2 - \frac{x^2}{4} + \frac{x^4}{48} \]

Thus, the Maclaurin series up to \( x^4 \) is:

\[ g(x) = \ln 2 - \frac{x^2}{4} + \frac{x^4}{48} \]

Detailed Answer

a) We are given:

\[ \frac{dy}{dx} = e^x - y \sin x \]

From this, we can tell that \( f'(0) = 1 \). Next, we take the second derivative of the right-hand side:

\[ \frac{d^2y}{dx^2} = e^x - \frac{dy}{dx}\sin{x} - y\cos{x} \]

Thus, \( f''(0) = -1 \). The Maclaurin series expansion of \( y \) up to \( x^2 \) is:

\[ y = 1 + x - \frac{x^2}{2} \]

b) Substituting \( x = 0.1 \):

\[ y \approx 1 + 0.1 - \frac{(0.1)^2}{2} \]

\[ y \approx 1 + 0.1 - \frac{0.01}{2}\]

\[ y \approx 1 + 0.1 - 0.005 \]

\[ y \approx 1 + \frac{11}{200} \]

Detailed Answer

a) The Maclaurin series is of the form:

\[ g(x) = \sum_{n=0}^{\infty} \frac{g^{(n)}(0)}{n!} x^n \]

Since \( g(0) = \ln(1 + 0) = \ln 1 = 0 \), the constant term is \( 0 \).

b) First, we compute the derivatives of \( g(x) \):

\[ g(x) = \ln(1 + x) \]

\[ g'(x) = \frac{1}{1 + x} \]

\[ g''(x) = -\frac{1}{(1 + x)^2} \]

\[ g'''(x) = \frac{2}{(1 + x)^3} \]

Evaluating at \( x = 0 \):

\[ g(0) = 0, \quad g'(0) = 1, \quad g''(0) = -1, \quad g'''(0) = 2 \]

Using the Maclaurin series formula:

\[ g(x) = g(0) + \frac{g'(0)}{1!} x + \frac{g''(0)}{2!} x^2 + \frac{g'''(0)}{3!} x^3 \]

Substituting the values:

\[ g(x) = 0 + x + \frac{-1}{2} x^2 + \frac{2}{6} x^3 \]

\[ g(x) = x - \frac{x^2}{2} + \frac{x^3}{3} \]

c) Since \( \ln 1.5 = g(0.5) \), we substitute \( x = 0.5 \):

\[ \ln 1.5 \approx 0.5 - \frac{(0.5)^2}{2} + \frac{(0.5)^3}{3} \]

\[ = 0.5 - \frac{0.25}{2} + \frac{0.125}{3} \]

Detailed Answer

a) Since \( g(x) = e^{\sin x} \), we differentiate:

\[ g'(x) = e^{\sin x} \cos x \]

\[ g''(x) = e^{\sin x} (\cos^2 x - \sin x) \]

Evaluating at \( x = 0 \):

\[ g(0) = e^0 = 1, \quad g'(0) = e^0 \cdot 1 = 1, \quad g''(0) = e^0 (1 - 0) = 1 \]

Using the Maclaurin expansion:

\[ g(x) = 1 + x + \frac{x^2}{2} \]

b) Differentiating again:

\[ g'''(x) = e^{\sin x} (\cos^3 x - 3 \sin x \cos x) \]

Evaluating at \( x = 0 \):

\[ g'''(0) = e^0 (1 - 0) = 1\]

Thus, the coefficient of \( x^3 \) is:

\[ \frac{g'''(0)}{3!} = \frac{1}{6} \]

c) Using:

\[ \cos x = 1 - \frac{x^2}{2} \]

\[ \ln(1 + \cos x) = \ln \left(1 + 1 - \frac{x^2}{2} \right) = \ln(2 - \frac{x^2}{2}) \]

Expanding \( \ln(1 + u) \approx u - \frac{u^2}{2} \) for small \( u \):

\[ \ln(1 + \cos x) = \ln 2 - \frac{x^2}{2} \]

d) We need to compute the following limit:

\[ \lim_{x \to 0} \frac{g(x) - 1}{x\ln(1 + \cos x)} = \lim_{x \to 0} \frac{x + \frac{x^2}{2}}{x\ln 2 - \frac{x^3}{2}} \]

Approximating for small \( x \):

\[ \approx \frac{x + \frac{x^2}{2}}{x\ln 2} \]

This can be written as:

\[ \frac{1}{\ln{2}}+\frac{x}{2\ln{2}} \dots \]

Thus, the limit is \( \frac{1}{\ln{2}} \).