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: $$ y'' = 0 + 0 + 0 + 6x + \cdots = \sum_{n=0}^{\infty} n(n-1)(n-2)x^{n-2} $$