Question

In the Taylor series expansion of function $f(x)=e^{x^{2}-x}$, coefficient of $x^{3}$ is

• A. $-\frac{7}{3}$
• B. -7
• C. $-\frac{7}{6}$
• D. $-\frac{7}{2}$

Hint:

For Taylor series questions, substituting a polynomial into a known standard series is almost always faster than taking repeated derivatives $f'''(0)/3!$.

Answer 1

The Correct Option is C

We will use the standard Maclaurin series expansion of the exponential function to find the specific coefficient.
Step 1: The expansion for $e^u$ is given by: $e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \frac{u^4}{4!} + \dots$. In our case, $u = x^2 - x$.
Step 2: $f(x) = 1 + (x^2 - x) + \frac{(x^2 - x)^2}{2} + \frac{(x^2 - x)^3}{6} + \dots$. We need to find all terms that produce an $x^3$ term upon expansion.
Step 3: 1) From the $(x^2 - x)$ term: No $x^3$ term is present. 2) From the $\frac{(x^2 - x)^2}{2}$ term: $\frac{1}{2}(x^4 - 2x^3 + x^2) \implies$ The $x^3$ term is $\frac{-2x^3}{2} = -x^3$. 3) From the $\frac{(x^2 - x)^3}{6}$ term: Using binomial expansion $(A-B)^3 = A^3 - 3A^2B + 3AB^2 - B^3$: $\frac{1}{6}((x^2)^3 - 3(x^2)^2x + 3x^2x^2 - x^3) = \frac{1}{6}(x^6 - 3x^5 + 3x^4 - x^3)$. The $x^3$ term is $-\frac{1}{6}x^3$.
Step 4: The total coefficient of $x^3$ is the sum of the coefficients identified in the previous step: Coefficient $= -1 + (-\frac{1}{6}) = -1 - \frac{1}{6} = -\frac{7}{6}$. This matches Option (3).