Given that $y = \sqrt{1+\ln(1+x)}$, find the exact range of values of $x$ for $y$ to be well defined. Show that $2y\,\dfrac{dy}{dx} = e^{-y^2}$. Hence, find the Maclaurin's series for $\sqrt{1+\ln(1+x)}$, up to and including the term in $x^2$. Verify that the same result is obtained using the standard series expansions given in the List of Formulae (MF15).

1. Set up the conditions for the natural logarithm and square root to be valid

For $y$ to be well defined, the argument of the logarithm must be strictly positive and the expression inside the square root must be non-negative:
$1+x > 0 \implies x > -1$
$1+\ln(1+x) \ge 0$

2. Solve the inequality to find the exact range of $x$

$\ln(1+x) \ge -1$
$1+x \ge e^{-1}$
$x \ge e^{-1} - 1$
Since $e^{-1} - 1 \approx -0.632 > -1$, this condition automatically satisfies $x > -1$.

1. Find y and dy/dx at x = 0 using the result of part (ii)

When $x=0$, $y=\sqrt{1+\ln 1}=1$.

From part (ii), $2y\dfrac{dy}{dx}=e^{1-y^2}$. Substituting $x=0,\;y=1$:
$2(1)\dfrac{dy}{dx}=e^{1-1}=1 \;\Rightarrow\; \left.\dfrac{dy}{dx}\right|_{x=0}=\tfrac{1}{2}.$

2. Differentiate the relation again to find d^2y/dx^2 at x = 0

Differentiate $2y\dfrac{dy}{dx}=e^{1-y^2}$ with respect to $x$:
$2\left(\dfrac{dy}{dx}\right)^2+2y\dfrac{d^2y}{dx^2}=e^{1-y^2}\left(-2y\dfrac{dy}{dx}\right).$
Since $e^{1-y^2}=2y\dfrac{dy}{dx}$, the right side $=-4y^2\left(\dfrac{dy}{dx}\right)^2$.

At $x=0$ ($y=1,\;\dfrac{dy}{dx}=\tfrac12$):
$2\left(\tfrac12\right)^2+2\dfrac{d^2y}{dx^2}=-4(1)\left(\tfrac12\right)^2$
$\tfrac12+2\dfrac{d^2y}{dx^2}=-1 \;\Rightarrow\; \left.\dfrac{d^2y}{dx^2}\right|_{x=0}=-\tfrac34.$

3. Assemble the Maclaurin series up to x^2

$y=y(0)+y'(0)\,x+\dfrac{y''(0)}{2!}x^2+\cdots$
$=1+\tfrac12 x+\dfrac{-\tfrac34}{2}x^2+\cdots=1+\tfrac12 x-\tfrac38 x^2+\cdots$

1. Write the standard series for $\ln(1+x)$

From MF15, $\ln(1+x) = x - \tfrac{1}{2}x^2 + \cdots$ so $1+\ln(1+x) = 1 + \left(x - \tfrac{1}{2}x^2\right) + \cdots$

2. Apply the binomial expansion $(1+u)^{1/2}$

Let $u = x - \tfrac{1}{2}x^2$. Then $\sqrt{1+\ln(1+x)} = (1+u)^{1/2} = 1 + \tfrac{1}{2}u - \tfrac{1}{8}u^2 + \cdots$ Since $u^2 = x^2 + \cdots$ (terms up to $x^2$), $= 1 + \tfrac{1}{2}\left(x - \tfrac{1}{2}x^2\right) - \tfrac{1}{8}x^2 + \cdots$

3. Collect terms and confirm agreement with (iii)

$= 1 + \tfrac{1}{2}x - \tfrac{1}{4}x^2 - \tfrac{1}{8}x^2 + \cdots = 1 + \tfrac{1}{2}x - \tfrac{3}{8}x^2 + \cdots$ This is identical to the series obtained in part (iii), as required.