Problem 39 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let $$ f(t)=\left\\{\begin{array}{ll} 0 & \text { if } t<0 \\ t^{4} & \text { if } t \geq 0 \end{array}\right. $$ Explain why \(f(t)\) cannot be represented by a Maclaurin series. Also show that, if \(g(t)\) gives the distance traveled by a car that is stationary for \(t<0\) and moving ahead for \(t \geq 0, g(t)\) cannot be represented by a Maclaurin series.

Step-by-Step Solution

Answer

Neither \( f(t) \) nor \( g(t) \) have derivatives of all orders at zero, so they can't be represented by a Maclaurin series.

Step 1: Understanding the Maclaurin Series

A Maclaurin series is a special type of Taylor series centered at zero. A function \( f(t) \) can be represented by a Maclaurin series if it has derivatives of all orders at zero and is equal to the series sum \( f(t) = \sum_{n=0}^{fty} \frac{f^{(n)}(0)}{n!} t^n \). This requires the function to be smooth and differentiable in a neighborhood around zero.

Step 2: Analyzing the Function At Zero

Examine the function \( f(t) \) at \( t=0 \). The function changes definition at this point, from 0 to \( t^4 \). The function is clearly defined at zero, where \( f(0) = 0^4 = 0 \).

Step 3: Checking for Continuity and Differentiability

The function \( f(t) \) is continuous at \( t=0 \), but not differentiable as it has a sudden change from a constant value (0 for \( t<0 \)) to a polynomial \( t^4 \) for \( t \geq 0 \). The derivative for \( t \geq 0 \) is \( f'(t) = 4t^3 \) and \( f'(0) = 0 \), but, the derivative from the left does not exist (since it was constant zero). Thus, \( f(t) \) is not smooth at \( t=0 \).

Step 4: Conclusion for f(t)

Since \( f(t) \) is not differentiable from all sides at \( t=0 \), it cannot have derivatives of all orders needed for a Maclaurin series, hence, it cannot be represented as one.

Step 5: Understanding g(t) Function

Similar to \( f(t) \), consider the nature of \( g(t) \) which represents distance. \( g(t) \) will have a form of essentially \( f(t) \) added with continuity conditions regarding car movement. It too follows the discontinuity in derivative as for \( t<0 \) the car is stationary, implying velocity (derivative) remains zero and for \( t \geq 0 \) starts moving.

Step 6: Analyzing g(t) at Zero

\( g(t) \) is stationary (0) for \( t<0 \) and moves for \( t\geq0 \). Its derivative from zero is not smooth across \( t=0 \) meaning it too cannot be expressed in terms of powers of \( t \) conflating negative and positive regimes into one series.

Step 7: Conclusion for g(t)

Similar to \( f(t) \), \( g(t) \) does not have derivatives of all orders at zero due to non-smoothness and thus, cannot be represented by a Maclaurin series.

Key Concepts

Differentiability

The concept of differentiability refers to the ability of a function to have a derivative at a particular point. Mathematically, a function is differentiable at a point if it has a well-defined tangent at that point. This means that there is a specific slope or a rate of change for the function as you look closely at that point.
For the function \( f(t) \), differentiability becomes an issue at \( t=0 \) because of its definition:

• When \( t<0 \), \( f(t) = 0 \), meaning the function is a flat line with a slope of zero.
• When \( t \geq 0 \), \( f(t) = t^4 \), the function follows a polynomial curve.

At \( t=0 \), the derivative from the left and the right differs because the flat line suddenly changes to a curve. The derivative from the right at \( t=0 \) is calculated from \( 4t^3 \), which is zero at \( t=0 \). However, from the left, there's no gradual change; it's just zero straight through. This absence of consistent slopes from both directions indicates \( f(t) \) isn't differentiable at this critical point.
As such, differentiability plays a crucial role here, because a Maclaurin series requires the function to be differentiable at and around zero to have derivatives of all orders.

Continuity

Continuity, as a concept, implies that a function does not have any abrupt changes at a particular point, which means you can draw the function without lifting your pencil. A function being continuous ensures that there are no holes, jumps, or vertical asymptotes at the point of interest. In simpler terms, a continuous function smoothly connects all its values.
For the function \( f(t) \), it is considered continuous at \( t=0 \) because:

• When \( t<0 \) it is zero, and right at \( t=0 \), \( f(0) = 0^4 = 0 \), ensuring no jump at this specific point.
• When \( t \geq 0 \), \( f(t) = t^4 \) continues the curve smoothly from zero onward.

Moreover, its continuity at \( t=0 \) indicates that it correctly passes through the point without skipping. This characteristic of being continuous, however, is not sufficient for a Maclaurin series representation because differentiability is also required.

Piecewise Functions

Piecewise functions, such as \( f(t) \) given in the exercise, are functions defined by different expressions over distinct intervals. They can pose challenges for forming series, such as the Maclaurin series, especially at the boundaries where they switch formulas.
In the case of \( f(t) \), it is defined as:

• \( f(t) = 0 \) for \( t<0 \), a constant expression.
• \( f(t) = t^4 \) for \( t \geq 0 \), a polynomial expression.

The change in definition at \( t=0 \) emphasizes those typical transitions inherent to piecewise functions. Notably, while the function addresses continuity, differentiability breaks down.
This complexity is why piecewise functions like \( f(t) \) cannot traditionally be expressed via a Maclaurin series unless smoothness between sections is guaranteed. Similarly, \( g(t) \) follows a piecewise pattern due to the car's behavior - stationary then moving, exhibiting non-smooth transitions, which further justifies the breakdown in using a Maclaurin series representation.