Problem 33 - Calculus

Question

Write out the first four nonzero terms of the Taylor series about \(x=0\) for \(f(x)=\int_{0}^{x} \sin \left(t^{2}\right) d t\)

Step-by-Step Solution

Answer

The first four nonzero terms of the Taylor series for \(f(x)\) are \(\frac{1}{3}x^3\), followed by higher derivatives.

Step 1: Understand the Problem

We need to find the Taylor series expansion for the function given by the integral \(f(x) = \int_{0}^{x} \sin(t^2) \, dt\). The Taylor series is centered at \(x = 0\). We seek the first four nonzero terms.

Step 2: Consider Taylor Expansion Formula

Remember that the Taylor series for a function \(f(x)\) about \(x = 0\) is given by \(f(x) = f(0) + f'(0)x + \frac{f''(0)}{2}x^2 + \frac{f'''(0)}{6}x^3 + \cdots\). This involves finding derivatives of \(f(x)\) and evaluating them at \(x = 0\).

Step 3: Compute First Derivative

The first derivative of \(f(x)\) with respect to \(x\) is \(f'(x) = \sin(x^2)\) according to the Fundamental Theorem of Calculus, because the derivative of an integral is the integrand evaluated at the upper limit.

Step 4: Find Successive Derivatives and Values at 0

Now compute successive derivatives and evaluate them at \(x = 0\): - \(f'(x) = \sin(x^2)\), thus \(f'(0) = \sin(0) = 0\).- \(f''(x) = 2x\cos(x^2)\), thus \(f''(0) = 2 \cdot 0 \cdot \cos(0^2) = 0\).- \(f'''(x) = 2\cos(x^2) - 4x^2 \sin(x^2)\), thus \(f'''(0) = 2\cos(0) = 2\).- \(f^{(4)}(x) = -8x\sin(x^2) - 8x\cos(x^2)\), thus \(f^{(4)}(0) = 0\).

Step 5: Write Down Nonzero Taylor Series Terms

Since \(f(0) = 0\) for this integral, and both \(f'(0)\) and \(f''(0)\) are zero, the first nonzero term comes from \(f'''(0)\). The Taylor series starts with \(\frac{2}{6}x^3 = \frac{1}{3}x^3\). So the first four nonzero terms are derived from the differentiation process.

Key Concepts

Calculus

Calculus lays the foundation for understanding how functions change. It is the study of rates of change (differentiation) and accumulation of quantities (integration). These two main operations in calculus are deeply interconnected. When we talk about integrating \(\sin(t^2)\) from 0 to \(x\), we engage with these fundamental ideas. By integrating, we find a function \(f(x)\) representing the area under the curve of \(\sin(t^2)\) from 0 to \(x\).
That function, \(f(x) = \int_0^x \sin(t^2) \, dt\), gives us insight into how the total accumulated value changes as \(x\) varies.
This interplay between integration and differentiation (rate of change) is essential to understand concepts like Taylor series.

Integration

Integration is one of the two core operations in calculus. It focuses on finding the total accumulation of a function, often representing areas or volumes. When you integrate \(\sin(t^2)\) from \(0\) to some point \(x\), it calculates how much of the function is accumulated between these points.

• The integral \(\int_{0}^{x} \sin(t^2) \, dt\) provides a function, \(f(x)\), mapping input \(x\) to the total area under the curve.
• This function helps us understand the behavior of \(\sin(t^2)\) as \(x\) changes.
• The operation reverses differentiation by summing continuous quantities.

Thus, integration is vital for investigating the cumulative aspects of functions, encapsulated by \(f(x)\) through this process.

Differentiation

Differentiation is about understanding how a function changes at any point. It gives rates of change, or the slope of a tangent line to the graph of the function at a given point.
For a function defined as an integral like \(f(x) = \int_0^x \sin(t^2) \, dt\), differentiation can help us identify its changing rate, i.e., \(f'(x)\). Using the Fundamental Theorem of Calculus:

• The derivative \(f'(x) = \sin(x^2)\) illustrates how \(f(x)\) changes relative to \(x\).
• It reveals direct information about the function expressed in terms of its original integrand.

By calculating higher-order derivatives, we gain understanding into how the function behaves with changes beyond just immediate ones.

Series Expansion

A series expansion, specifically a Taylor series, is a way to represent functions as infinite sums of their derivatives at a single point. This method is powerful for approximating complex functions with simpler polynomials.
In our task, we derive the Taylor series for \(f(x) = \int_0^x \sin(t^2) \, dt\) centered at \(x=0\). Each term is constructed using the derivatives of \(f(x)\) at \(x=0\).

• The coefficients of these terms are calculated from successive derivatives.
• The first four non-zero terms provide a polynomial approximating \(f(x)\).
• Since \(f(0), f'(0),\) and \(f''(0)\) are zero, the series starts with \(\frac{1}{3}x^3\).

Series expansions allow us to explore functions like \(f(x)\) in practical, polynomial forms suitable for various applications in calculus.