Question

Use the remainder term to find a bound on the error in the following approximations on the given interval. Error bounds are not unique. $$e^{x}=1+x+\frac{x^{2}}{2} \text {on } \left[-\frac{1}{2}, \frac{1}{2}\right]$$

Short answer

Answer: The bound on the error for the given approximation on the interval \(\left[-\frac{1}{2}, \frac{1}{2}\right]\) is \(E \leq \frac{e^{\frac{1}{2}}}{48}\).

Step-by-step solution

Find the \((n+1)^\text{th}\) derivative of the function

Since the given function is \(f(x) = e^x\), we need to find its \((n+1)^\text{th}\) derivative. The derivatives of \(e^x\) are: $$f^{(1)}(x) = e^x$$ $$f^{(2)}(x) = e^x$$ $$f^{(3)}(x) = e^x$$ Since the third and all higher derivatives of \(e^x\) are also equal to \(e^x\), we have: $$f^{(n+1)}(x) = f^{(3)}(x) = e^x$$

Determine the maximum value of the \((n+1)^\text{th}\) derivative on the given interval

We need to find the maximum value of \(f^{(3)}(x) = e^x\) on the interval \(\left[-\frac{1}{2}, \frac{1}{2}\right]\). Since \(e^x\) is an increasing function, the maximum value will occur at the right endpoint of the interval: $$\max_{-\frac{1}{2}\leq x \leq \frac{1}{2}} e^x = e^{\frac{1}{2}}$$

Apply the remainder term formula for the error bound

Now that we have the maximum value of the \((n+1)^\text{th}\) derivative on the given interval, we can use the remainder term formula to find a bound on the error: $$R_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!}x^{n+1}$$ Plugging in the values we have found, we get: $$R_2(x) = \frac{e^{\frac{1}{2}}}{(2+1)!}x^{2+1}$$ $$R_2(x) = \frac{e^{\frac{1}{2}}}{6}x^3$$ To find the maximum error, we need to find the maximum value of the absolute value of the remainder term on the given interval: $$E = \max_{-\frac{1}{2}\leq x \leq \frac{1}{2}} \left|\frac{e^{\frac{1}{2}}}{6}x^3\right|$$ Since the largest magnitude of \(x\) on the interval \([-\frac{1}{2}, \frac{1}{2}]\) is \(\frac{1}{2}\), the maximum error will occur at this point: $$E = \frac{e^{\frac{1}{2}}}{6}\left(\frac{1}{2}\right)^3 = \frac{e^{\frac{1}{2}}}{48}$$ So, the bound on the error for the given approximation on the interval \(\left[-\frac{1}{2}, \frac{1}{2}\right]\) is: $$E \leq \frac{e^{\frac{1}{2}}}{48}$$

Key concepts

Taylor Series

The Taylor series is a representation of a function as an infinite sum of terms, each of which is derived from the function's derivatives at a single point. It is named after the mathematician Brook Taylor who introduced this concept. If you've got a smooth function, you can use the Taylor series to approximate it around a reference point, often chosen as zero (which would make it a Maclaurin series— a special case of the Taylor series).
To construct a Taylor series for the function, we start with an infinite sum:
\[\begin{equation} f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots \end{equation}\]
The 'a' in the parentheses is the point around which the function is approximated, while the derivatives are evaluated at this point. Each additional term in the series includes a higher-order derivative divided by the factorial of the order. This sequence continues infinitely, offering a more precise approximation the further you go.

Remainder Term

The remainder term in the context of the Taylor series quantifies the error difference between the actual value of the function and the Taylor polynomial's estimated value. It essentially tells us how good our approximation is.
For a Taylor series of order 'n', the remainder term, often denoted as 'R_n(x)', can be interpreted as the error introduced when truncating the series after the nth-term. There are different forms of the remainder term, one of them being the Lagrange form:
\[\begin{equation} R_n(x) = \frac{f^{(n+1)}(\text{xi})}{(n+1)!}(x-a)^{n+1} \end{equation}\]
Here, the \text{xi} is some value in the interval between 'a' and 'x', and it is often not known exactly. The point \text{xi} is kind of a mystery guest in our equation, representing any possible value that could deliver the very worst-case scenario of the error.

Exponential Function Approximation

The exponential function, denoted as \(e^x\), is one of the most important functions in mathematics, especially because of its unique property where its derivative is equal to the function itself. Approximating this function using a Taylor series becomes incredibly useful in various areas such as physics, engineering, and finance.
To approximate \(e^x\), we can truncate its Taylor series to a finite number of terms. The original exercise provided an approximation valid within the interval of \(\left[-\frac{1}{2}, \frac{1}{2}\right]\):
\[\begin{equation} e^{x}=1+x+\frac{x^{2}}{2} \end{equation}\]
It’s a second-order polynomial, meaning it only takes into account the function and its first two derivatives. Such a simplification is helpful for calculations, especially when the exact value isn't practically necessary or computing the value is too complex.

Error Estimation

Error estimation is the process of assessing the inaccuracy in a given approximation. When we truncate a Taylor series, as in the exercise, it's important to know how far off our approximation might be from the true value—we call this the 'error bound'.
In the exercise, we used the remainder term of the Taylor series to determine the maximum possible error in approximating the exponential function on a given interval. By calculating the maximum value of the remainder term within this interval, we have:
\[\begin{equation} E \leq \frac{e^\frac{1}{2}}{48} \end{equation}\]
This inequality shows us that any x-value within the interval \(\left[-\frac{1}{2}, \frac{1}{2}\right]\) will yield an approximation of \(e^x\) with an error less than or equal to \(\frac{e^\frac{1}{2}}{48}\). Understanding error estimation enables us to use these approximations responsibly, ensuring we recognize the limits within which they are sufficiently accurate.