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In a multichoice online test that I did the other day, I was required to select the Maclaurin series for $e^{\tan(x)}$. It was necessary for me to find the first four terms in order to establish which answer was correct. In the end of year exam, I will have reference to a Useful Information booklet (this contains a generalized Taylor polynomial approximation and Maclaurin series of $e^x$, $(1+x)^n$, $\sin(x)$, $\cos(x)$ and $\ln(1+x)$), and no calculator - hence throughout all of of my work, including online tests (which do contribute to my grade), I choose only to work with this resource, as preparation for this exam.

As my approach for this problem, I used the generalized Taylor polynomial approximation to find the Maclaurin series for $\tan(x)$ and substituted this series in place of $x$ into the given Maclaurin series for $e^x$, and double-checked my answer once I had finished every other question by using the generalized Taylor polynomial approximation to find the Maclaurin series for $e^{\tan(x)}$. Obviously, as you can imagine, both of these methods were very time consuming (especially when you consider that the other nineteen questions in the test collectively took me less than ten minutes to answer).

I'm probably missing a simple concept here. Can you please help me to establish a more elegant approach to this problem?

The choices I was given were as follows:

Options

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Why the negative vote? –  clookid Sep 1 '12 at 23:54
    
What were the choices you were given? Probably all but the correct one were easily ruled out. –  Jonathan Sep 2 '12 at 0:06
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I was given six choices. Three were easily ruled out. But the remaining three required that I find at least the fourth term. I've added the choices above. –  clookid Sep 2 '12 at 0:17
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5 Answers 5

up vote 10 down vote accepted

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?

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Brilliant method. Thank you very much. –  clookid Sep 2 '12 at 0:48
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Based on the choices given in your picture:

Choices given

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 $e^x$.

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Thanks - another great approach! I'm a little bit uncertain as to how you're able to compute that for values of $x$ just greater than zero $e^{tan(x)} > e^x$, in your head, though. –  clookid Sep 2 '12 at 1:03
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$\tan(x)>x$ for $x>0$ (and $x<\pi/2$ of course) –  Jonathan Sep 2 '12 at 1:04
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Without doubt the fastest method of all.. –  nbubis Sep 2 '12 at 1:08
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This is much better than the other answers. On a multiple choice exam you do not want to take five minutes to do some elaborate calculation when you can quickly eliminate a bunch of wrong choices. –  MJD Sep 2 '12 at 2:50
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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 $x$.

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That's a very elegant method of finding the series for $tan(x)$! Thank you. –  clookid Sep 2 '12 at 1:04
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See this article: Exponential formula

If you know the expansion for $f(x)$ in powers of $x$, what is the expansion of $e^{f(x)}$? That's what the exponential formula tells you.

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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 $$ e^{\tan x} \approx \frac 1{0!} + \frac 1{1!} x + \frac{1}{2!}x^2 + \frac3{3!} x^3= 1 + x + \frac 12 x^2 + \frac 12 x^3. $$

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This is the approach that I took to verify my answer. I'm more interested in an approach that will not require me to compute the third derivative of $e^{tan(x)}$! –  clookid Sep 2 '12 at 0:20
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You can simplify computation of derivatives if you keep in mind your goal is evaluation at $x=0$. Also, since $\tan(x)$ is an odd function you know its even-order derivatives are zero. Wise use of these facts can speed the calculus. –  James S. Cook Sep 2 '12 at 0:30
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