Question

How do you find the power representation for the function $f\left( x \right) = \dfrac{{1 + x}}{{1 - x}}$?

Answer 1

As we have to find the power representation of the function. First, find the expansion of $\dfrac{1}{{1 - x}}$ by the formula ${\left( {1 - x} \right)^{ - 1}} = 1 + x + {x^2} + \ldots$. After that find the expansion of $\dfrac{x}{{1 - x}}$ by multiplying the formula by $x$ on both sides. Then, add both the expansion to get the desired result.

Complete step-by-step solution:
It is given in the question that we have to find the power representation for the function $f\left( x \right) = \dfrac{{1 + x}}{{1 - x}}$ which means we have to write the expansion of the function.
Many mathematical functions may be expressed in the form of power series. The power series is a series of powers or the sum of a sequence. The number of elements can be finite or infinite in the set. As a function of any vector (say x), a power series can be called.
In mathematical analysis, power series are useful where they arise as Taylor's series of infinitely differentiable functions. We should know how to write the power series/McLaurin series of a form function to solve this question.
As we know that the expansion of ${\left( {1 - x} \right)^{ - 1}}$ is given by,
$\Rightarrow {\left( {1 - x} \right)^{ - 1}} = 1 + x + {x^2} + \ldots$
As we can write,
$\Rightarrow {\left( {1 - x} \right)^{ - 1}} = \dfrac{1}{{1 - x}}$
So, the above expansion can be written as,
$\Rightarrow \dfrac{1}{{1 - x}} = 1 + x + {x^2} + \ldots$ ….. (1)
Now multiply above expansion by $x$ on both sides to get the expansion of $\dfrac{x}{{1 - x}}$,
$\Rightarrow x \times \dfrac{1}{{1 - x}} = x\left( {1 + x + {x^2} + \ldots } \right)$
Multiply the terms,
$\Rightarrow \dfrac{x}{{1 - x}} = x + {x^2} + {x^3} + \ldots$ ….. (2)
As we know,
$\Rightarrow \dfrac{{1 + x}}{{1 - x}} = \dfrac{1}{{1 - x}} + \dfrac{x}{{1 - x}}$
Substitute the values from equation (1) and (2),
$\Rightarrow \dfrac{{1 + x}}{{1 - x}} = \left( {1 + x + {x^2} + \ldots } \right) + \left( {x + {x^2} + {x^3} + \ldots } \right)$
Add the terms,
$\Rightarrow \dfrac{{1 + x}}{{1 - x}} = 1 + 2x + 2{x^2} + 2{x^3} + \ldots$
Hence, the power representation for the function $f\left( x \right) = \dfrac{{1 + x}}{{1 - x}}$ is $1 + 2x + 2{x^2} + 2{x^3} + \ldots$.

Note: When we add two series that consist of the infinite terms then we add coefficient terms with the same power terms. Also, note that that expansion of ${\left( {1 + x} \right)^n}$, where n is negative or fraction is given as:
${\left( {1 + x} \right)^n} = 1 + nx + \dfrac{{n\left( {n - 1} \right)}}{2}{x^2} + \ldots$.