In the Taylor’s series expansion of ex about x = 2, the coefficient of (x – 2)4 is
e2/4!
Taylor's Series Expansion Basics
The Taylor's series expansion is a way to represent a function as an infinite sum of terms, where each term is calculated from the function's derivatives at a single point. This point is often called the "center" of the expansion.
For a function \(f(x)\) that is infinitely differentiable at a point \(x = a\), its Taylor's series expansion around \(x = a\) is given by the formula:
$$ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x - a)^n $$
This can be written out as:
$$ f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \frac{f^{(4)}(a)}{4!}(x - a)^4 + \dots $$
In this question, we are asked to find the coefficient of \((x - 2)^4\) in the Taylor's series expansion of \(e^x\) about \(x = 2\).
Function and Center Identification
Based on the question, we have:
- The function is \(f(x) = e^x\).
- The expansion is about the point \(x = a = 2\).
- We need the coefficient of the term where the power of \((x - a)\) is 4, which means \(n = 4\).
From the general Taylor's series formula, the coefficient of the \((x - a)^n\) term is given by \(\frac{f^{(n)}(a)}{n!}\). For our specific case, the coefficient of \((x - 2)^4\) will be \(\frac{f^{(4)}(2)}{4!}\).
Derivatives of the Exponential Function
To find the required coefficient, we first need to calculate the derivatives of the function \(f(x) = e^x\) up to the 4th order.
| Order of Derivative | Derivative \(f^{(n)}(x)\) |
|---|---|
| \(f(x)\) (0th derivative) | \(e^x\) |
| \(f'(x)\) (1st derivative) | \(e^x\) |
| \(f''(x)\) (2nd derivative) | \(e^x\) |
| \(f'''(x)\) (3rd derivative) | \(e^x\) |
| \(f^{(4)}(x)\) (4th derivative) | \(e^x\) |
It is clear that for the exponential function \(e^x\), every derivative is simply \(e^x\).
Coefficient Calculation for (x – 2)4
Now we need to evaluate the 4th derivative at the center of expansion, \(x = a = 2\):
$$ f^{(4)}(2) = e^2 $$
Finally, we can find the coefficient of \((x - 2)^4\) by plugging this value into the coefficient formula:
$$ \text{Coefficient of } (x - 2)^4 = \frac{f^{(4)}(2)}{4!} = \frac{e^2}{4!} $$
Result Summary
The coefficient of \((x - 2)^4\) in the Taylor's series expansion of \(e^x\) about \(x = 2\) is \(\frac{e^2}{4!}\).
Comparing this result with the given options:
- Option 1: \(2^4/4!\)
- Option 2: \(1/4!\)
- Option 3: \(e^2/4!\)
- Option 4: \(e^4/4!\)
The calculated coefficient matches Option 3.
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