Simple Maclaurin Series $e^{\tan(x)}$

Solution 1:

Try the following. If you only want the first four terms then you can compute everything $\bmod x^4$. Then

\begin{eqnarray*} \tan x &=& \frac{\sin x}{\cos x} \\ &\equiv& \frac{x - \frac{x^3}{6}}{1 - \frac{x^2}{2}} \bmod x^4 \\ &\equiv& \left( x - \frac{x^3}{6} \right) \left( 1 + \frac{x^2}{2} \right) \bmod x^4 \\ &\equiv& x + \frac{x^3}{3} \bmod x^4. \end{eqnarray*}

Then

\begin{eqnarray*} e^{\tan x} &\equiv& e^x e^{\frac{x^3}{3} } \bmod x^4 \\ &\equiv& \left( 1 + x + \frac{x^2}{2} + \frac{x^3}{6} \right) \left( 1 + \frac{x^3}{3} \right) \bmod x^4 \\ &\equiv& 1 + x + \frac{x^2}{2} + \frac{x^3}{2} \bmod x^4. \end{eqnarray*}

This took a little under 5 minutes on computer, and by hand it probably would have been a little faster. Is that fast enough?

Solution 2:

Based on the choices given in your picture:

choices a, b, e, f can be immediately ruled out: choice a because you recognize that it's the expansion of $\exp(x)$, choice e because $\tan x\sim x$ and $\exp(x)\sim 1+x$ tell us the $x$ term should be there. Choices b and f are joke answers.

Choice d is wrong because, for $x$ just bigger than zero, $\exp(\tan x)>e^x$ while the expansion in choice d is smaller than the expansion of $e^x$.

Solution 3:

If you are already given the expansions for $\sin(x),\cos(x)$ and $e^x$, then: $$\tan{x}\simeq a_1x +a_3x^3+\cdots =\frac{x - x^3/6 + \cdots}{1 - x^2/2+\cdots}$$ (Since $\tan x$ is odd). Multiply both sides, and you can easily find the values of $a_0$: $$a_1 = 1,-1/2 + a_3 = -1/6 \rightarrow a_3=1/3$$ Note that you don't actually have to preform the full multiplication. So now: $$e^{\tan x} = 1 + (x + x^3/3) + (x + x^3/3)^2/2 + (x + x^3/3)^3/6 + \cdots $$ $$= 1 + x + x^2/2 + x^3/2 + \cdots$$ Where again you don't need to carry out the full multiplication - only the terms that you see contribute to the lower powers in $x$.

Solution 4:

You can simply use the definition of the Taylor series:
$$ f(x) \approx \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!} (x-a)^n . $$ To use this, you first need to find the derivatives of the function $f(x) = e^{\tan x}$, evaluated at wherever you want to center the series. These first few derivatives are \begin{align*} f(x) &= e^{\tan x} \\ f'(x) &= e^{\tan x} \sec^2 x \\ f''(x) &= e^{\tan x}(\sec^4 x + \sec^2 x \tan x) \\ f'''(x) &= e^{\tan x}(\sec^6 x + 2\sec^4 x + 6\sec^4 x \tan x + 4 \sec^2 x \tan^2 x) \end{align*} and so we if we want the first few terms of a MacLaurin series, we evaluate these derivatives at 0 to get $f(0) = 1$, $f'(0) = 1$, $f''(0) = 1$ and $f'''(x) = 3$. Then the fourth Taylor polynomial is given by $$ e^{\tan x} \approx \frac 1{0!} + \frac 1{1!} x + \frac{1}{2!}x^2 + \frac3{3!} x^3= 1 + x + \frac 12 x^2 + \frac 12 x^3. $$