Question

The Taylor series for an infinitely differentiable function \(f\) at \(x=a\) is given by $$ T(x)=\sum_{k=0}^{\infty} \frac{f^{(k)}(a)}{k !}(x-a)^{k} $$ where \(f^{(k)}\) is the \(k\) th derivative of \(f(\) see \([5, \S 8.8])\). Let \(f(x)=e^{x^{2}}\). (a) Find (by hand) the first three nonzero terms of the Taylor series for \(f\) at \(x=0\). (b) Check your answer to part (a) by defining \(f\) as a symbolic function and using the taylor function: \(\gg\) syms \(x\) \(\gg f=\exp \left(x^{-} 2\right)\) \(\gg \mathrm{T} 5=\operatorname{tay} \operatorname{lor}(\mathrm{f}, \mathrm{x},\) 'order' \(, 5)\) (c) Plot the function and the Taylor polynomial on the same axes. (d) Use the Taylor polynomial to approximate \(\int_{0}^{1} e^{x^{2}} d x\).

Short answer

The first three nonzero terms are \(1 + x^2\). The integral approximation is \(\frac{4}{3}\).

Step-by-step solution

Find Derivatives

To find the first three non-zero terms of the Taylor series for the function \( f(x) = e^{x^2} \) at \( x=0 \), we first calculate the derivatives.1. \( f(x) = e^{x^2} \). 2. \( f'(x) = 2x e^{x^2} \). 3. \( f''(x) = (4x^2+2)e^{x^2} \). 4. \( f'''(x) = (8x^3 + 12x) e^{x^2} \).

Evaluate at x=0

Evaluate each derivative at \( x = 0 \) to find non-zero terms.1. \( f(0) = e^{0} = 1 \).2. \( f'(0) = 2 \cdot 0 \cdot e^{0} = 0 \).3. \( f''(0) = (4 \cdot 0^2 + 2) e^{0} = 2 \). 4. \( f'''(0) = (8 \cdot 0^3 + 12 \cdot 0) e^{0} = 0 \).

Use Taylor Series Formula

The Taylor series up to the first three non-zero terms is formulated as follows: \[ T(x) = \frac{f(0)}{0!}(x-0)^0 + \frac{f''(0)}{2!}(x-0)^2 + \cdots \]For this exercise, it becomes:\[ T(x) = 1 + \frac{2}{2}(x^2) = 1 + x^2 \]

Verify Using Symbolic Math

In symbolic math, define function and calculate series. First, we declare \(x\) as a symbolic variable and \(f = \exp(x^2)\). Then, we use MATLAB's function to find the Taylor expansion: \[ \text{syms x} \] \[ \text{f = exp}(x \hat{} 2) \] \[ \text{T5 = taylor}(f, x, 'order', 5) \] Which should output \(1 + x^2\).

Graph Function and Polynomial

Plot \( f(x) = e^{x^2} \) and its approximated Taylor polynomial \( T(x) = 1 + x^2 \) on the same set of axes to visually compare them.

Approximate the Integral

Use the polynomial to approximate the integral: \[\int_{0}^{1} e^{x^2} \, dx \approx \int_{0}^{1} (1 + x^2) \, dx = \left[ x + \frac{x^3}{3} \right]_{0}^{1} = 1 + \frac{1}{3} = \frac{4}{3}.\]

Key concepts

Symbolic Math

Symbolic math allows us to work with mathematical expressions in a symbolic form rather than just as numeric values. This is especially useful when dealing with complex functions like the Taylor series. In symbolic math, variables are treated as symbols which can be manipulated according to algebraic rules.
When performing symbolic operations, one can differentiate or integrate functions without assigning specific numerical values to the variables. It’s useful for theoretical computations and achieving exact results. In the context of finding a Taylor series, symbolic math helps by allowing us to represent functions and their derivatives easily.
Using tools like MATLAB or SymPy in Python, we can declare a function symbolically and then apply operations like differentiation or Taylor expansions automatically, as shown in the exercise solution. This greatly simplifies the process, making complex symbolic calculations accessible to anyone.

Derivatives

Derivatives play a crucial role in the formation of Taylor series. They represent the rate at which a function changes. For the Taylor series computation, derivatives of a function at a specific point (usually denoted as point "a") determine the coefficients of the series.
When dealing with the Taylor series of a function like \( e^{x^2} \), we calculate several derivatives to unpack the behavior of the function around the point of expansion, usually around \( x = 0 \) for simplicity. In the exercise, we calculated the first few derivatives of \( e^{x^2} \) and evaluated them at 0:

• \( f(x) = e^{x^2} \)
• \( f'(x) = 2x e^{x^2} \)
• \( f''(x) = (4x^2+2)e^{x^2} \)
• \( f'''(x) = (8x^3 + 12x) e^{x^2} \)

By evaluating these derivatives, we identified the coefficients for the Taylor series. Understanding derivatives deeply can help you comprehend how the behavior of the parent function influences the series.

Function Approximation

Function approximation is a fundamental aspect of using Taylor series. The idea is to approximate a complex function with a simpler polynomial that behaves similarly around a specific point.
The technique involves using derivatives of the function at this specific point, which leads to a polynomial that resembles the original function over a certain interval. In the exercise, the function \( e^{x^2} \) at \( x=0 \) is approximated by a polynomial derived from its Taylor series:

• Main function: \( f(x) = e^{x^2} \)
• Approximating polynomial: \( T(x) = 1 + x^2 \)

The more terms we include from the Taylor series, the closer the approximation becomes, though sometimes even just a few terms can provide significant insights into the function's nature. This is why Taylor series are highly valuable in both pure and applied mathematics.

Integral Approximation

Integral approximation using Taylor series allows us to estimate the value of an integral that might otherwise be challenging to compute directly.
In the context of our exercise, we used the approximate polynomial \( T(x) = 1 + x^2 \) to estimate the integral of \( e^{x^2} \) over the interval \([0, 1]\). Calculating the integral of a polynomial is often simpler than dealing with the original function:

• Approximated integral: \( \int_{0}^{1} e^{x^2} \, dx \approx \int_{0}^{1} (1 + x^2) \, dx \)
• Result: \( \left[ x + \frac{x^3}{3} \right]_{0}^{1} = 1 + \frac{1}{3} = \frac{4}{3} \)

This approach greatly simplifies complex problems. Taylor series make otherwise difficult integration tasks approachable, especially in engineering and physics contexts, where exact solutions might be unnecessary or impossible to find.