Problem 75 - Single Variable Calculus

Question

Let \(f\) be the function defined by $$ f(x)=\left\\{\begin{array}{ll} e^{-1 / x^{2}} & \text { if } x \neq 0 \\ 0 & \text { if } x=0 \end{array}\right. $$ Show that \(f\) cannot be represented by a Maclaurin series.

Step-by-Step Solution

Answer

The function f(x) cannot be represented by a Maclaurin series because its derivatives at x = 0 are equal to 0, and the Maclaurin series does not converge to the function \(f(x) = e^{-1/x^{2}}\) for x ≠ 0.

Step 1: Compute the derivatives of f(x) for x ≠ 0

To begin, let's find the derivatives of the function f(x) for x ≠ 0 since it is the value not equal to 0 of x for which the function is differentiable. We will repeatedly apply the product rule and chain rule to differentiate f(x) = \(e^{-1/x^{2}}\): \(f'(x) = \frac{d}{dx} \left( e^{-1/x^{2}} \right) = e^{-1/x^{2}} \frac{d}{dx} \left( -\frac{1}{x^{2}} \right) = 2x e^{-1/x^{2}}\) Now, we will compute the second derivative: \(f''(x) = \frac{d}{dx} \left( 2x e^{-1/x^{2}} \right) = 2 e^{-1/x^{2}} + 2x(-2x^{-3}) e^{-1/x^{2}} = 2 e^{-1/x^{2}} (1 - 4x^{4})\) We can continue this process, but this should be enough to see a pattern.

Step 2: Evaluate the derivatives at x = 0

Now, we need to evaluate the derivatives of f(x) at x = 0: \(f(0) = 0\) \(f'(0) = 2 \cdot 0 e^{-1/0^{2}} = 0\) \(f''(0) = 2 \cdot e^{-1/0^{2}} (1 - 4 \cdot 0^{4}) = 0\) It's clear that the function and all its derivatives at x = 0 are equal to 0.

Step 3: Examine the convergence of the Maclaurin series

From the above analysis, we see that the function and its derivatives at x = 0 are always equal to 0. The Maclaurin series of f(x) would be given as: \(f(x) = 0 + 0 \cdot x + 0 \cdot \frac{x^{2}}{2!} + 0 \cdot \frac{x^{3}}{3!} + \cdots = 0\) However, when x ≠ 0, the function f(x) = \(e^{-1/x^{2}}\) is not equal to 0. Therefore, the Maclaurin series does not converge to the function for x ≠ 0.

Step 4: Conclusion

Based on the above analysis, we have shown that the function f(x) cannot be represented by a Maclaurin series because its derivatives at x = 0 are equal to 0, and the Maclaurin series does not converge to the function for x ≠ 0.

Key Concepts

Derivatives

In calculus, derivatives measure how a function changes as its input changes. For the function \(f(x) = e^{-1/x^2}\) when \(x eq 0\), we can use rules like the product and chain rules to find derivatives.
These help us understand the behavior of the function around specific points. When calculating derivatives, you start with the first derivative and can continue to higher orders.

• The first derivative \(f'(x)\) gives us the rate of change or the slope of the function.
• The second derivative \(f''(x)\) provides insight into the concavity or curvature.

For this function, all derivatives at \(x = 0\) turn out to be zero, which significantly impacts the ability to form a Maclaurin series.

Convergence

Convergence refers to whether a sequence or series approaches a specific value. In the context of a Maclaurin series, we want the series to converge to the original function.
For \(f(x)\), we have:

• All derivatives at \(x = 0\) equal 0.
• The series becomes \(0 + 0 \cdot x + 0 \cdot \frac{x^2}{2!} + \ldots = 0\), which is just zero.

This series doesn't match \(f(x) = e^{-1/x^2}\) when \(x eq 0\). Therefore, the series does not converge to the function, showing a mismatch outside \(x = 0\).

Function Representation

A function can often be represented by a series expansion like a Maclaurin series, which is a Taylor series centered at zero. This series allows a polynomial representation of a function.
The function \(f(x)\) is defined in a piecewise manner:

• \(e^{-1/x^2}\) for \(x eq 0\)
• 0 for \(x = 0\)

A Maclaurin series representation requires matching the behavior of the function at \(x = 0\), but here it fails because it always returns zero, not matching the exponential component when \(x\) is not zero.

Calculus

Calculus is the mathematical study of change, using concepts like derivatives and integrals. In studying \(f(x) = e^{-1/x^2}\), calculus helps assess continuity, differentiability, and series representation.
Calculus allows us to:

• Find derivatives to understand the function's growth and decay.
• Determine if series representations like Maclaurin series are possible based on derivative behavior.

Through these tools, we conclude that our function cannot be adequately captured by a Maclaurin series.

Differentiation Rules

Differentiation rules are essential in finding the derivatives of functions. Using the product and chain rules, we can differentiate complex expressions like \(e^{-1/x^2}\).
Here's how we apply them:

• The chain rule helps differentiate compositions of functions.
• The product rule is used when functions multiply.

By applying these rules, we systematically differentiate \(f(x)\) to show that all derivatives vanish at \(x = 0\). This results in the breakdown of representation by the Maclaurin series, as every term becomes zero.