Q-28E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Question: Show that,$28. \sum _{n=0}^{\infty }{a}_n{x}^{n+1}+\sum _{n=1}^{\infty }n{b}_n{x}^{n-1}=b_1+\sum _{n-1}^{\infty }[2a_{n-1}+(n+1\left)b_{n+1}\right]x^n$

Step-by-Step Solution

Answer

We showed that 2 .$\sum _{n=0}^{\infty }{a}_n{x}^{n+1}+\sum _{n=1}^{\infty }n{b}_n{x}^{n-1}=b_1+\sum _{n-1}^{\infty }[2a_{n-1}+(n+1\left)b_{n+1}\right]x^n$

Step 1: Power series

A power series is an infinite series of the form,
$\sum _{n=0}^{\infty }{a}_n{(x-c)}^n=a_0+ a_1(x-c) +a_2(x-c{)}^2+.....$
Where,a_n represents the coefficient term of the nth term and c is a constant.

Step 2: Changing the index of the first term

We have to show that,

2.$\sum _{n=0}^{\infty }{a}_n{x}^{n+1}+\sum _{n=1}^{\infty }n{b}_n{x}^{n-1}=b_1+\sum _{n-1}^{\infty }[2a_{n-1}+(n+1\left)b_{n+1}\right]x^n$

Simplifying the L.H.S expression,

L.H.S =2.$\sum _{n=0}^{\infty }{a}_n{x}^{n+1}+\sum _{n-1}^{\infty }n{b}_n{x}^{n-1}$

Now changing the index of the first term, let,

n + 1 = k

      n = k -1

Then,

$$\begin{gathered} 2 \sum _{n=0}^{\infty }{a}_n{x}^{n+1}=2 \sum _{k-1=0}^{\infty }{a}_{k-1}{x}^k \\ =2 \sum _{k=0}^{\infty }{a}_{k-1}{x}^k \end{gathered}$$

The index is a dummy variable, so we can replace ^k with ⁿ , the first term of the L.H.S becomes$, 2 \sum _{k=0}^{\infty }{a}_{k-1}{x}^k$

Step 3: Changing the index of the second term

Now changing the index of the second term, let,

n - 1 = k

      n = k + 1

Then,

$$\begin{gathered} \sum _{n-1}^{\infty }n{b}_n{x}^{n-1}=\sum _{k+1=1}^{\infty }(k+1)b_{k+1}{x}^k \\ =\sum _{k=0}^{\infty }(k+1)b_{k+1}{x}^k \end{gathered}$$

The index is a dummy variable, so we can replace ^k with ⁿ , the first term of the L.H.S becomes $=\sum _{k=0}^{\infty }(k+1)b_{k+1}{x}^k$

The second expression in L.H.S, starts with index 0 , in order to combine both the expressions, we will take the second expression from n=1

 ,$$\begin{gathered} 2. \sum _{n=0}^{\infty }{a}_n{x}^{n+1}+\sum _{n=1}^{\infty }n{b}_n{x}^{n-1}=2. \sum _{n=1}^{\infty }{a}_{n-1}{x}^n+b_1+ \sum _{n=0}^{\infty }(n+1)b_{n+1}{x}^n \\ =b_1+ \sum _{n=1}^{\infty }\left[2a_{n-1}+(n+1)b_{n+1}\right]x^n \end{gathered}$$

which is equal to the R.H.S of the given statement.

 

Therefore, by simplifying the L.H.S of the expression, we can prove that both the expressions are the same.