91Ó°ÊÓ

A Taylor polynomial is a finite sum of terms derived from the Taylor series of a function. It provides an approximation of a function around a specific point. In mathematical terms, the Taylor series is an infinite sum of terms calculated from the values of the function's derivatives at a single point. The Taylor polynomial, which is a truncated version of the Taylor series, helps in estimating the value of the function near the point around which it is expanded. For instance, if we consider the function \(e^{x}\), we can use its Taylor polynomial to approximate values like \(e^{0.01}\) accurately.

The general form of a Taylor series at point \(a\) is given by:
\[ T_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \frac{f^{(n)}(a)}{n!}(x-a)^n \]

• \(T_n(x)\) is the Taylor polynomial of degree \(n\).
• \(f^{(n)}(a)\) denotes the \(n\)-th derivative of the function evaluated at point \(a\).
• The factorial \(n!\) serves to normalize the term of the series.

Derivatives are essential to the construction of Taylor polynomials. A derivative measures how a function changes as its input changes. For Taylor polynomials, we need the function's value and its derivatives up to the desired degree.

For the function \(f(x) = e^x\), an interesting property is that all derivatives of \(e^x\) are itself. This simplifies our calculations significantly. Here's how they look:

• \(f'(x) = e^x\)
• \(f''(x) = e^x\)
• \(f'''(x) = e^x\)
• and so on...

At \(x = 0\), this means:

• \(f(0) = 1\)
• \(f'(0) = 1\)
• \(f''(0) = 1\)
• \(f'''(0) = 1\)
• \(f^{(4)}(0) = 1\)

This repetitive pattern allows us to substitute these values directly into the Taylor series formula for easy computation.

The exponential function \(e^x\) is one of the most crucial functions in mathematics, given its unique properties and applications across different fields. One standout characteristic is that the derivative of \(e^x\) is \(e^x\), making it an ideal candidate for Taylor series expansions.

To form a Taylor polynomial for \(e^x\) around \(x = 0\), follow these steps:

• Identify the value of \(e^x\) and its higher-order derivatives at \(x = 0\).
• Since \(e^0 = 1\), \(f(0)\), \(f'(0)\), etc., all equal 1.
• Substitute these values into the Taylor series up to the desired polynomial degree.
• For a fourth-degree Taylor polynomial, substitute into:
  \[ T_4(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} \]

This results in
\[ T_4(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} \]

Estimation using this polynomial:

• To estimate \(e^{0.01}\), substitute \(x = 0.01\) into the polynomial.
• Perform the arithmetic operations to get \(e^{0.01} \approx 1.01005017\).

This estimation is impressively accurate, showcasing the power of Taylor polynomials in approximating functions like \(e^x\).