Question

For what values of \(p\) does the Taylor series for \(f(x)=(1+x)^{p}\) centered at 0 terminate?

Short answer

The Taylor series for the function \(f(x)=(1+x)^{p}\) centered at 0 terminates for integer values of \(p\) such that \(p\in\{0, 1, 2, ... n\}\).

Step-by-step solution

Start with the given function

We are given the function \(f(x) = (1+x)^{p}\).

Find the n-th order derivative of the function

To find the general term of the Taylor series, we need to find the n-th order derivative of the function. For the given function, let's find the first few derivatives: - \(f(x) = (1+x)^{p}\), - \(f'(x) = p(1+x)^{p-1}\), - \(f''(x) = p(p-1)(1+x)^{p-2}\), - \(f^{(3)}(x) = p(p-1)(p-2)(1+x)^{p-3}\), As we can notice, the process can be generalized as: - \(f^{(n)}(x) = p(p-1)(p-2)...(p-n+1)(1+x)^{p-n}\).

Evaluate the n-th derivative at x=0

To obtain the Taylor series of the function centered at 0, we need to evaluate the n-th derivative of the function at x=0: - \(f^{(n)}(0) = p(p-1)(p-2)...(p-n+1)(1+0)^{p-n}\), - \(f^{(n)}(0) = p(p-1)(p-2)...(p-n+1)\).

Find the general term of the Taylor series

The general term of the Taylor series consists of the coefficients \(f^{(n)}(0)/n!\) and the term \(x^n\). We now have: \(\frac{f^{(n)}(0)}{n!}x^n = \frac{p(p-1)(p-2)...(p-n+1)}{n!}x^n\).

Determine when the Taylor series terminates

Since the Taylor series terminates when its coefficients \(f^{(n)}(0)/n!\) become zero, we need to find for what values of \(p\) the coefficients become zero for some n. For \(p=0,1,2,...,n\), if \(n\leq p\), the factor \((p-n+1)\) in the coefficient will be zero. Then, the coefficient will be zero. In conclusion, the Taylor series for the function \(f(x)=(1+x)^{p}\) centered at 0 terminates for integer values of \(p\) such that \(p\in\{0, 1, 2, ... n\}\).

Key concepts

Derivatives

Derivatives are a core concept in calculus, used to understand how functions change. For a given function, the derivative at a point gives the slope of the tangent line to the curve of the function at that point. In the context of our original exercise, we're looking at derivatives of a specific function:

• The function is given as \(f(x) = (1+x)^p\).
• To construct the Taylor series, we need many levels of derivatives: first, second, third, and so on.
• The pattern for increasing derivatives involves multiplying the previous derivative by a factor related to \(p\).

Understanding the derivative process helps build the Taylor series around a center, such as 0, by finding out how those derivatives behave and contribute to the series terms.

General Term

The general term of a Taylor series is crucial because it tells us what each term in the series looks like. For the function \(f(x) = (1+x)^p\), each term is derived using the derivatives of the function evaluated at the center. Here's how it works:

• Each term in the Taylor series is given by \(\frac{f^{(n)}(0)}{n!}x^n\).
• This means we use the \(n\)-th derivative at \(x = 0\) divided by \(n!\), the factorial of \(n\), to determine the coefficient of \(x^n\).
• The general pattern for the derivatives of our function gives us the format \(p(p-1)...(p-n+1)\), showing how each step involves reducing \(p\) by additional integers as \(n\) increases.

This pattern helps us recognize when the series might stop, leading to the concept of series termination.

Series Termination

A Taylor series can terminate - or, in simpler terms, stop growing in terms of its complexity and number of non-zero terms. This happens when the coefficients of further terms become zero. For the exercise function, this occurs:

• The Taylor series will contain zero for higher powers if the coefficient sequence \(p(p-1)...(p-n+1)\) equals zero.
• This wipes out the term in the expansion, effectively 'terminating' the series at that point.
• This occurs specifically when \(p\) allows one of the factors in \(p(p-1)...\) to become zero, indicating that the sequence's value is zero, rendering the series finite.

Understanding when and how this termination happens is key in identifying what values of \(p\) result in a finite series.

Integer Values

In the context of this problem, integer values of \(p\) play an essential role in determining when the Taylor series will terminate. Here is why integers matter:

• When \(p\) is an integer, the expression \(p(p-1)...(p-n+1)\) will eventually contain a zero factor, causing the series to stop forming new terms.
• The series effectively 'dies out' after a finite number of terms because integer math inevitably generates a zero at some point in multiplying down from \(p\).
• In simpler terms, integer values of \(p\) ensure a predetermined number of non-zero terms, confirming the termination of the series. These integer breaks directly link to factorial reductions in the series' formula.

Thus, integer values are critical in this context as they define the boundary where the Taylor series simplifies itself into a polynomial.