GATE 2018 EC Q.24: solution and answer

GATE 2018 · ECQ.24 · 1 mark · NAT

Taylor series expansion of $f (x) = \int_{0}^{x} e^{- (\frac{t ^{2}}{2})} d t$ around $x = 0$ has the form

f (x) = a_{0} + a_{1} x + a_{2} x^{2} + \dots

The coefficient $a_{2}$ (correct to two decimal places) is equal to ________.

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Answer and solution

ANSWER

$a_{2} = 0.00$ . The integrand $e^{- t^{2} /2}$ is even, so $f (x) = \int_{0}^{x} e^{- t^{2} /2} d t$ is odd and its Maclaurin series has no $x^{2}$ term: $f (x) = x - \frac{x ^{3}}{6} + \frac{x ^{5}}{40} - \dots$ .

Official answer: -0.01 to 0.01Source: Official answer key, GATE 2018 EC (IIT Guwahati)

The Taylor coefficient about $x = 0$ is $a_{n} = \frac{f ^{(n)} ( 0 )}{n !}$ , so $a_{2} = \frac{f ^{''} ( 0 )}{2 !} = \frac{f ^{''} ( 0 )}{2}$ .
By the fundamental theorem of calculus (Leibniz rule with a variable upper limit and a constant lower limit), $f^{'} (x) = e^{- x^{2} /2}$ .
Differentiate again with the chain rule: $f^{''} (x) = \frac{d}{d x} e^{- x^{2} /2} = e^{- x^{2} /2} \cdot \frac{d}{d x} (- \frac{x ^{2}}{2}) = - x e^{- x^{2} /2}$ .
At $x = 0$ : $f^{''} (0) = - 0 \cdot e^{0} = 0$ , so $a_{2} = \frac{0}{2} = 0$ .
Check by integrating the series term by term. With $e^{u} = 1 + u + \frac{u ^{2}}{2 !} + \dots$ and $u = - \frac{t ^{2}}{2}$ : $e^{- t^{2} /2} = 1 - \frac{t ^{2}}{2} + \frac{t ^{4}}{8} - \dots$ , so $f (x) = \int_{0}^{x} (1 - \frac{t ^{2}}{2} + \frac{t ^{4}}{8} - \dots) d t = x - \frac{x ^{3}}{6} + \frac{x ^{5}}{40} - \dots$ Only odd powers appear, because the integrand is even, which makes $f$ an odd function.
So $a_{0} = 0$ , $a_{1} = 1$ and $a_{2} = 0$ . The answer is $a_{2} = 0.00$ .

KEY IDEA

You never need to evaluate the integral. Differentiate under the variable upper limit to get $f^{'} (x) = e^{- x^{2} /2}$ , then use $a_{2} = f^{''} (0) /2$ . You can also see it without calculating: the integral of an even function from $0$ to $x$ is an odd function, and an odd function's Maclaurin series has only odd powers. So every even-indexed coefficient, including $a_{2}$ , is zero.

WATCH OUT FOR

Stopping at $f^{'} (0) = 1$ and reporting it as $a_{2}$ is the usual slip, and so is forgetting to divide by $2!$ . That second error does no harm here only because $f^{''} (0) = 0$ . Another mistake is to read the $t^{2} /2$ in the exponent as a coefficient in the answer, writing $a_{2} = - \frac{1}{2}$ . That $- \frac{1}{2}$ is the $t^{2}$ coefficient of the integrand, and integrating moves it to the $x^{3}$ term of $f$ , where it becomes $- \frac{1}{6}$ .

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GATE 2018 EC Q.24: solution and answer

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