Question

a. Differentiate the Taylor series about 0 for the following functions. b. Identify the function represented by the differentiated series. c. Give the interval of convergence of the power series for the derivative. $$f(x)=\sqrt{1+x}$$

Short answer

The differentiated Taylor series represents the function \(f'(x) = \frac{1}{1 + \frac{x}{2}}\), and its interval of convergence is \((-2, 2)\).

Step-by-step solution

Find the Taylor series for the given function

Recall that the Taylor series of a function \(f(x)\) about \(0\) is given by: $$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n \ = \ f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3+ \cdots $$ where \(f^{(n)}(0)\) denotes the \(n\)-th derivative of the function evaluated at \(0\). First, let's find the derivatives of the function \(f(x) = \sqrt{1+x}\) at \(x=0\): $$f'(x) = \frac{1}{2(1+x)^{\frac{1}{2}}} \Rightarrow f'(0)=\frac{1}{2}$$ $$f''(x) = -\frac{1}{4(1+x)^{\frac{3}{2}}} \Rightarrow f''(0)=-\frac{1}{4}$$ $$f'''(x) = \frac{3}{8(1+x)^{\frac{5}{2}}} \Rightarrow f'''(0)=\frac{3}{8}$$ Notice a pattern between the derivatives and their values at \(0\). We can generalize and find that for a positive integer \(n\): $$f^{(n)}(0) =\frac{(-1)^{n+1}(n-1)!}{2^n}$$ Thus, the Taylor series for \(f(x) = \sqrt{1+x}\) about \(0\) is: $$f(x) = 1 + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(n-1)!}{2^n n!}x^n$$

Differentiate the Taylor series

Differentiate both sides of the Taylor series with respect to \(x\): $$f'(x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(n-1)!}{2^n n!} \cdot n x^{n-1} = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(n-1)!}{2^n (n-1)!}x^{n-1} = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{2^n}x^{n-1}$$

Identify the function represented by the differentiated series

Observe that the differentiated series is a geometric series: $$f'(x) = 1 - \frac{x}{2} + \frac{x^2}{4} - \frac{x^3}{8} + \cdots$$ with a ratio of \(-\frac{x}{2}\) and an initial term of \(1\). Therefore, we can find the sum of this geometric series. Recall that the sum of a geometric series with ratio \(r\) and first term \(a\) is given by \(\frac{a}{1 - r}\), provided that \(|r| < 1\). As a result, we have: $$f'(x) = \frac{1}{1 + \frac{x}{2}}$$

Determine the interval of convergence for the power series of the derivative

The convergence of the power series for the derivative depends on the geometric series ratio \(|-\frac{x}{2}| < 1\). Solving this inequality for \(x\) yields the interval of convergence: $$|-x/2|<1 \Rightarrow -2 < x < 2$$ Hence, the interval of convergence for the power series of the derivative is \((-2, 2)\).

Key concepts

Power Series

A power series is a series of terms in the form \( a_n x^n \), where each term consists of a coefficient \( a_n \) and a variable \( x \) raised to a power \( n \), starting from zero and going to infinity. It's written as:\[S(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + \cdots = \sum_{n=0}^{\infty} a_nx^n\] Power series are incredibly useful in calculus because they allow us to express complex functions in a simpler form that's easier to work with. To find a power series representation of a function, you start by calculating its derivatives at a given point and substituting these into the power series formula. In the given exercise, we found the Taylor series of \(f(x) = \sqrt{1+x}\), a specific type of power series, about "0" (also known as a Maclaurin series). This process involves finding successive derivatives of the original function evaluated at zero, thus obtaining the coefficients \( a_n \). Simplifying and summing these terms provides the power series for further analysis or manipulation.

Interval of Convergence

The interval of convergence is the range of \( x \) values for which the power series converges or is valid. For a power series \( \sum_{n=0}^\infty a_n (x - c)^n \), this interval determines where the series represents a real number. To find this interval, you typically investigate the series' behavior using the ratio test or other methods to determine where the series sums to a finite value. In the context of the exercise, once we differentiate the Taylor series, we identify it as a geometric series with a ratio of \(-\frac{x}{2}\). Therefore, we calculate the interval of convergence using the condition for a geometric series to converge, \(|r| < 1\). This gives us \(|-\frac{x}{2}| < 1\), leading to the interval \(-2 < x < 2\). This means that within this interval, the differentiated series accurately represents the function.

Geometric Series

A geometric series is a series where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. The simplest form of a geometric series is:\[S = a + ar + ar^2 + ar^3 + \cdots = \sum_{n=0}^{\infty} ar^n\] This series converges when the absolute value of the common ratio \( |r| \) is less than 1, and its sum can be calculated using the formula:\[\text{Sum} = \frac{a}{1 - r}\] In our exercise, by differentiating the Taylor series of \(\sqrt{1+x}\), we resulted in a geometric series with an initial term of "1" and a common ratio of \(-\frac{x}{2}\). Identifying this power series as a geometric series allows us to utilize the convergence criteria and sum formula to find the corresponding function, which turned out to be \(f'(x) = \frac{1}{1 + \frac{x}{2}}\). Understanding geometric series is vital as it provides a straightforward approach to sum these series provided the interval of convergence is adhered to.