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The question is to find the value of the Maclaurin series expansion of $e^{\arctan x}$ up to (but not including) $\mathcal{O}(n^5)$.

I tried using the Maclaurin series for $e^{u}$ then subbing $\arctan x$ into it, and I got $$\sum_{n=0}^{\infty } \frac{(\arctan x)^{n}}{n!}$$

Is this the correct approach to solving this problem or am I missing something? Cause if so I'm not sure when to stop to reach $\mathcal{O}(n^5)$

Gaurang Tandon
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Calvin Woo
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  • just calculate the deriviates at $0$. no need to complicate. – dezdichado Feb 11 '18 at 04:00
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    I presume you mean $O(x^5)$ in your question, since there is no $n$ there. Do you know the Maclaurin series for $\arctan x$? If so, then you can simply plug it in to the series for $e^t$; this is known as series composition. (Convince yourself that you'll only need the terms up to $n=5$ in both series, and that you can throw away any higher-order terms as they arise) – Steven Stadnicki Feb 11 '18 at 04:05
  • Possible duplicate of Multiplying Taylor series and composition – Gaurang Tandon Feb 11 '18 at 04:37
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    @GaurangTandon : How is this a duplicate of that? That was about pointwise multplication of power series, which becomes Cauchy multiplication of the sequences of coefficients. But in this present problem, one is composing functions, not multiplying them. – Michael Hardy Feb 11 '18 at 06:00
  • @MichaelHardy I was looking at the second part of the answer by Asaf Karagila, which began with "Composition". OP's question is a composition of $e^x$ and $\arctan x$, hence I flagged as duplicate. – Gaurang Tandon Feb 11 '18 at 06:02

3 Answers3

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https://en.wikipedia.org/wiki/Exponential_formula

\begin{align} \arctan x & = x - \frac{x^3} 3 + \frac{x^5} 5 - \frac{x^7} 7 + \cdots \\[10pt] & = a_1 x + a_2 \frac{x^2} 2 + a_3 \frac{x^3} 6 + a_4 \frac{x^4}{24} + \cdots \\[10pt] a_1 & = 1 \\ a_2 & = 0 \\ a_3 & = -1/3 \\ a_4 & = 0 \\ a_5 & = +1/5 \\ & \,\,\,\vdots \\[10pt] e^{\arctan x} & = 1 + b_1 x + b_2 \frac{x^2} 2 + b_3 \frac{x^3} 6 + b_4 \frac{x^4} {24} + \cdots \\[10pt] b_1 & = a_1 \\ b_2 & = a_1^2 + a_2 \\ b_3 & = a_1^3 + 3a_1 a_2 + a_3 \\ b_4 & = a_1^4 + 4a_1 a_3 + 3a_2^2 + 6a_1^2 a_2 + a_4 \\ & \text{etc.} \end{align} The pattern is this:

The coefficient of $a_2^2$ is $3$ because there are three ways to partition a set of four objects into two sets of two: $$ ab/cd, \qquad ac/bd, \qquad ad/bc $$ The coefficient of $a_1^2 a_2$ is $6$ because there are six ways to partition a set of four objects into two sets of one and a set of two: $$ a/b/cd, \qquad a/c/bd, \qquad a/d/bc, \qquad b/c/ad, \qquad b/d/ac, \qquad c/d/ab $$ And so on.

Thus $$ e^{\arctan x} = 1 + x + x^2 + \frac{x^3} 9 - \frac{x^4}{72} + \cdots $$ You should probably check my arithmetic.

Next we have $$ b_5 = a_1^5 + 10a_1^3 a_2 + 15a_1 a^2_2 + 10a_1^2 a_3 + 10a_2 a_3 + 5a_1 a_4 + a_5. $$

Michael Hardy
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Composition of series is a very interesting problem.

Suppose $$A=e^{f(x)} \implies \log(A)=f(x)$$ and develop $f(x)$ as $$f(x)=f(0)+x f'(0)+\frac{1}{2} x^2 f''(0)+\frac{1}{6} f^{(3)}(0) x^3+\frac{1}{24} f^{(4)}(0) x^4+O\left(x^5\right)$$ Now, use $A=e^{\log(A)}$ and use the expansion of the exponential function. You should get $$A=e^{f(0)}+e^{f(0)} f'(0)\,x+\frac{1}{2} e^{f(0)} \left(f''(0)+f'(0)^2\right) x^2+\frac{1}{6} e^{f(0)} \left(f^{(3)}(0)+f'(0)^3+3 f'(0) f''(0)\right)x^3 +\frac{1}{24} e^{f(0)} \left(f^{(4)}(0)+3 f''(0)^2+f'(0)^4+4 f^{(3)}(0) f'(0)+6 f'(0)^2 f''(0)\right) x^4+O\left(x^5\right)$$ which, for your case, simplify a lot since $f(0)=0$, $f'(0)=1$, $f''(0)=0$, $f^{(3)}(0)=-2$, $f^{(4)}(0)=0$.

In the case of an odd function, this will simplify as $$A=1+ f'(0)\,x+\frac{1}{2} f'(0)^2x^2+\frac{1}{6} \left(f^{(3)}(0)+f'(0)^3\right)x^3+\frac{1}{24} \left(f'(0)^4+4 f^{(3)}(0) f'(0)\right)x^4+O\left(x^5\right)$$

Claude Leibovici
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It is enough to differentiate $f(x)=e^{\arctan x}$ five times, by exploiting $g'(x)=g(x)\cdot\frac{d}{dx}\log g(x)$:

$$ f'(x)=\frac{1}{1+x^2}f(x),\qquad f''(x)=\frac{1-2x}{(1+x^2)^2}f(x),\qquad f'''(x)=\frac{-1-6x+6x^2}{(1+x^2)^3}f(x),$$ $$ f^{IV}(x) = \frac{-7+12x+36x^2-24x^3}{(1+x^2)^4}f(x),\quad f^{V}(x)=\frac{5+140 x-120 x^2-240 x^3+120 x^4}{(1+x^2)^5}f(x) $$ since $f(0)=1$, by evaluating these derivatives at $x=0$ we get $$ e^{\arctan x} = 1+x+\frac{x^2}{2}-\frac{x^3}{6}-\frac{7 x^4}{24}+\frac{x^5}{24}+o(x^5) $$ in a neighbourhood of the origin.

Jack D'Aurizio
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