To find the coefficients [tex]\( c_0, c_1, c_2, c_3, c_4 \)[/tex] of the power series [tex]\(\sum_{n=0}^{\infty} c_n x^n\)[/tex] for [tex]\(\frac{E(x)}{1-x}\)[/tex], we proceed as follows:
1. Define the power series [tex]\( E(x) \)[/tex]:
[tex]\[
E(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots
\][/tex]
2. Define the power series for [tex]\( \frac{1}{1-x} \)[/tex], which is a geometric series:
[tex]\[
\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + x^4 + \cdots
\][/tex]
3. Multiply the two power series:
[tex]\[
\frac{E(x)}{1-x} = \left(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \right) \left(1 + x + x^2 + x^3 + \cdots \right)
\][/tex]
4. Find the coefficients [tex]\( c_n \)[/tex] by multiplying the series term-by-term and collecting like powers of [tex]\( x \)[/tex]. Let's go through the terms one by one until [tex]\( x^4 \)[/tex]:
- Coefficient [tex]\( c_0 \)[/tex]:
[tex]\[
c_0 = 1 \cdot 1 = 1
\][/tex]
- Coefficient [tex]\( c_1 \)[/tex]:
[tex]\[
c_1 = 1 \cdot x + x \cdot 1 = 1 + 1 = 2
\][/tex]
- Coefficient [tex]\( c_2 \)[/tex]:
[tex]\[
c_2 = 1 \cdot x^2 + x \cdot x + \frac{x^2}{2!} \cdot 1 = 1 + 1 + \frac{1}{2} = 2.5
\][/tex]
- Coefficient [tex]\( c_3 \)[/tex]:
[tex]\[
c_3 = 1 \cdot x^3 + x \cdot x^2 + \frac{x^2}{2!} \cdot x + \frac{x^3}{3!} \cdot 1 = 1 + 1 + \frac{1}{2} + \frac{1}{6} = \frac{8}{3} \approx 2.6666666666666665
\][/tex]
- Coefficient [tex]\( c_4 \)[/tex]:
[tex]\[
c_4 = 1 \cdot x^4 + x \cdot x^3 + \frac{x^2}{2!} \cdot x^2 + \frac{x^3}{3!} \cdot x + \frac{x^4}{4!} \cdot 1 = 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} = 2.7083
\][/tex]
Therefore, the coefficients are:
[tex]\[
c_0 = 1, \quad c_1 = 2, \quad c_2 = 2.5, \quad c_3 = \frac{8}{3} \approx 2.6666666666666665, \quad c_4 = 2.7083.
\][/tex]
So, [tex]\( \frac{E(x)}{1-x} = c_0 + c_1 x + c_2 x^2 + c_3 x^3 + c_4 x^4 + \cdots \)[/tex] with the coefficients [tex]\( c_0, c_1, c_2, c_3, \)[/tex] and [tex]\( c_4 \)[/tex] as calculated above.
