Question

Given that \(y=\sum_{n=0}^{\infty} n x^{n},\) compute \(y^{\prime}\) and \(y^{\prime \prime}\) and write out the first four terms of each series as well as the coefficient of \(x^{n}\) in the general term.

Short answer

Question: Write down the first four terms and the general term for the first and second derivatives of the power series \(y=\sum_{n=0}^{\infty} nx^n\). Answer: For the first derivative \(y'\): The first four terms are \(0 + 0 + 2x + 6x^2\) and the general term is given by \(y'= \sum_{n=0}^{\infty} n(n-1)x^{n-1}\). For the second derivative \(y''\): The first four terms are \(0 + 0 + 0 + 6x\) and the general term is given by \(y''= \sum_{n=0}^{\infty} n(n-1)(n-2)x^{n-2}\).

Step-by-step solution

Find the general formula for the first derivative \(y'\)

To find the derivative of a power series, we can take the derivative of each term in the series and sum them up. The derivative of the general term \(n x^n\) is obtained using the power rule for differentiation: \((n x^n)' = n(n-1)x^{n-1}\). The series \(y'\) will be given by: $$ y'=\sum_{n=0}^{\infty}n(n-1)x^{n-1} $$

Find the first four terms of the first derivative \(y'\)

Now, let's calculate the first four terms of the series \(y'\). When \(n=0\): $$ 0(0-1)x^{-1} = 0 $$ When \(n=1\): $$ 1(1-1)x^{1-1} = 0 $$ When \(n=2\): $$ 2(2-1)x^{2-1} = 2x $$ When \(n=3\): $$ 3(3-1)x^{3-1} = 6x^2 $$ So, the first four terms of the series are: $$ y' = 0 + 0 + 2x + 6x^2 + \cdots $$ The general term for the first derivative is: $$ y'= \sum_{n=0}^{\infty} n(n-1)x^{n-1} $$

Find the general formula for the second derivative \(y''\)

We will apply the same method we used in step 1 to the first derivative: take the derivative of each term in the series and sum them up. The derivative of the general term \(n(n-1)x^{n-1}\) is \((n(n-1)x^{n-1})' = n(n-1)(n-2)x^{n-2}\). The series \(y''\) will be given by: $$ y''=\sum_{n=0}^{\infty}n(n-1)(n-2)x^{n-2} $$

Find the first four terms of the second derivative \(y''\)

Now let's calculate the first four terms of the series \(y''\). When \(n=0\): $$ 0(0-1)(0-2)x^{-2} = 0 $$ When \(n=1\): $$ 1(1-1)(1-2)x^{1-2} = 0 $$ When \(n=2\): $$ 2(2-1)(2-2)x^{2-2} = 0 $$ When \(n=3\): $$ 3(3-1)(3-2)x^{3-2} = 6x $$ So, the first four terms of the series are: $$ y'' = 0 + 0 + 0 + 6x + \cdots $$ The general term for the second derivative is: $$ y''= \sum_{n=0}^{\infty} n(n-1)(n-2)x^{n-2} $$ In conclusion, the first derivative \(y'\) is given by: $$ y' = 0 + 0 + 2x + 6x^2 + \cdots = \sum_{n=0}^{\infty} n(n-1)x^{n-1} $$ And the second derivative \(y''\) is given by: $$ y'' = 0 + 0 + 0 + 6x + \cdots = \sum_{n=0}^{\infty} n(n-1)(n-2)x^{n-2} $$

Key concepts

First Derivative

Power Series differentiation is a systematic technique used to find the derivatives of a function expressed as a power series. Let's dive into finding the first derivative of the given power series. The function is given as:\[y = \sum_{n=0}^{\infty} n x^n.\]To obtain the first derivative of this series, we use the power rule for differentiation, \((x^n)' = na^{n-1}\). Therefore, by differentiating each term of the series, we get its derivative as:\[y' = \sum_{n=0}^{\infty} n(n-1)x^{n-1}.\]Here is a breakdown of calculating each term for better understanding:

• When \(n=0\): The term becomes \(0(0-1)x^{-1} = 0\).
• When \(n=1\): You get \(1(1-1)x^{0} = 0\).
• When \(n=2\): The result is \(2(2-1)x^{1} = 2x\).
• When \(n=3\): It simplifies to \(3(3-1)x^{2} = 6x^2\).

Thus, the first four terms of the derivative are:\[y' = 0 + 0 + 2x + 6x^2 + \cdots.\]

Second Derivative

Now, we will move forward to compute the second derivative of our power series. Similar to finding the first derivative, we will differentiate the first derivative expression:\[y' = \sum_{n=0}^{\infty} n(n-1)x^{n-1}.\]Using the power rule applied once more:\[(x^{n-1})' = (n-1)x^{n-2},\]we derive the second derivative of each term:\(y'' = \sum_{n=0}^{\infty} n(n-1)(n-2)x^{n-2}.\)Breaking it further down by computing each of the initial terms:

• For \(n=0\): The outcome is \(0(0-1)(0-2)x^{-2} = 0\).
• At \(n=1\): You end up with \(1(0)(-1)x^{-1} = 0\).
• When \(n=2\): It results in \(2(1)(0)x^{0} = 0\).
• At \(n=3\): The value becomes \(3(2)(1)x^{1} = 6x\).

Therefore, the first four terms of the second derivative series are:\[y'' = 0 + 0 + 0 + 6x + \cdots.\]

General Term Coefficient

Identifying the coefficient of \(x^n\) in the general term of a power series is critical for understanding its behavior. This coefficient guides how each term in the series contributes to the function.For the original series \(y = \sum_{n=0}^{\infty} n x^n\), the coefficient of \(x^n\) is simply \(n\). This highlights that as \(n\) increases, each term contributes more significantly in the series.When we differentiate to obtain \(y'\), the coefficient for each \(x^{n-1}\) becomes \(n(n-1)\). Essentially, you multiply the previous coefficient \(n\) by \(n-1\), indicating a faster increase in contribution from each term as \(n\) grows.Similarly, for the second derivative \(y''\), the coefficient evolves to \(n(n-1)(n-2)\) for \(x^{n-2}\). Now, each term in the series contributes even more because of this factor multiplication with \(n-2\). Consistently, as differentiation progresses, the impact of the higher powers of \(n\) becomes more prominent in the coefficient, emphasizing the increased weight of higher-order terms in the function’s derivative. Understanding these coefficients helps predict how quickly the series terms escalate as \(n\) rises.