Derive Taylor Expansion of $f(x)$ Around x=0 Up to x^2

[Guest2]

Consider the function $f(x) =\sqrt{1 + \sin(x)}$. Derive the first few terms in the Taylor expansion of $f(x)$ around $x = 0$, up to and including terms of order $x^2$. Give an explicit formula for the remainder term.

I can do the first part, but how do I find the remainder term?

 

[jvicens]

To find the remainder term, we can use the Lagrange form of the remainder in Taylor's theorem. This states that the remainder term $R_n(x)$ for a function $f(x)$ with a Taylor polynomial of degree $n$ centered at $x=a$ is given by:

$R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$

where $c$ is some number between $a$ and $x$. In this case, since we are finding the Taylor expansion around $x=0$, our formula becomes:

$R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}$

To find the remainder term for $f(x) = \sqrt{1 + \sin(x)}$, we first need to find the derivatives of this function. We can use the chain rule to find the derivatives of the inner function $1+\sin(x)$, which are:

$f'(x) = \frac{\cos(x)}{2\sqrt{1+\sin(x)}}$

$f''(x) = -\frac{\sin(x) + 2\cos^2(x)}{4(1+\sin(x))^{\frac{3}{2}}}$

$f'''(x) = \frac{3\sin^2(x) - 6\sin(x)\cos^2(x) - 2\cos^4(x)}{8(1+\sin(x))^{\frac{5}{2}}}$

Now, we can plug these derivatives into our formula for the remainder term, and since we are only interested in terms up to $x^2$, we will use $n=2$. This gives us:

$R_2(x) = \frac{f^{(3)}(c)}{3!}x^3 = \frac{\frac{3\sin^2(c) - 6\sin(c)\cos^2(c) - 2\cos^4(c)}{8(1+\sin(c))^{\frac{5}{2}}}}{3!}x^3$

We can simplify this to get our final formula for the remainder term:

$R_2(x) = \frac{\sin^2(c) - 2\sin(c)\cos^2(c) - \