
How do you find the Maclaurin series for \[\dfrac{1}{{1 - x}}\]?
Answer
559.8k+ views
Hint: In the given question, we have been given to solve for the Maclaurin series for a given polynomial expression. For that, we make the given polynomial such that we bring all the terms to one side, so that on one side of equality there is only the constant one. Then we start to write the multiplier for the polynomial on the other side of equality so that it is equal to “\[1\]”. Then we generalize the series so as to get the required Maclaurin series for the given polynomial.
Complete step-by-step answer:
The given series is \[\dfrac{1}{{1 - x}}\].
The most effective way to find the Maclaurin series is to start writing the multipliers for \[\left( {1 - x} \right)\] which make it equal to \[1\],
$\Rightarrow$ \[1 = \left( {1 - x} \right)\left( {...} \right.\]
First, we multiply \[1\], so that the \[1\] in \[\left( {1 - x} \right)\] gives \[1\],
$\Rightarrow$ \[1 = \left( {1 - x} \right)\left( {1 + ...} \right.\]
Now, when we multiplied \[1\], the other term of the polynomial \[\left( { - x} \right)\] of \[\left( {1 - x} \right)\] gave \[\left( { - x} \right)\], so now we are going to have to multiply with \[\left( x \right)\] so that the \[\left( { - x} \right)\] term cancels out,
$\Rightarrow$ \[1 = \left( {1 - x} \right)\left( {1 + x + ...} \right.\]
Then it gives \[\left( { - {x^2}} \right)\] and we multiply it with its multiplicative inverse \[\left( {{x^2}} \right)\] and this system goes on forever, hence, we have,
$\Rightarrow$ \[1 = \left( {1 - x} \right)\left( {1 + x + {x^2} + {x^3} + ...} \right.\]
So, \[\dfrac{1}{{1 - x}} = 1 + x + {x^2} + {x^3} + ... = \sum\limits_{k = 0}^\infty {{x^k}} \]
Note: In the given question, we had to find the Maclaurin series for a polynomial. We did that by keeping the constant \[1\] on one side of equality and the rest of the terms of the polynomial on the other. Then we started writing the multiplier for the polynomial so that it became equal to \[1\]. Then we generalized the series so as to get the required Maclaurin series for the given polynomial. So, it is really important that we know the formulae and where, when and how to use them so that we can get the correct result.
Complete step-by-step answer:
The given series is \[\dfrac{1}{{1 - x}}\].
The most effective way to find the Maclaurin series is to start writing the multipliers for \[\left( {1 - x} \right)\] which make it equal to \[1\],
$\Rightarrow$ \[1 = \left( {1 - x} \right)\left( {...} \right.\]
First, we multiply \[1\], so that the \[1\] in \[\left( {1 - x} \right)\] gives \[1\],
$\Rightarrow$ \[1 = \left( {1 - x} \right)\left( {1 + ...} \right.\]
Now, when we multiplied \[1\], the other term of the polynomial \[\left( { - x} \right)\] of \[\left( {1 - x} \right)\] gave \[\left( { - x} \right)\], so now we are going to have to multiply with \[\left( x \right)\] so that the \[\left( { - x} \right)\] term cancels out,
$\Rightarrow$ \[1 = \left( {1 - x} \right)\left( {1 + x + ...} \right.\]
Then it gives \[\left( { - {x^2}} \right)\] and we multiply it with its multiplicative inverse \[\left( {{x^2}} \right)\] and this system goes on forever, hence, we have,
$\Rightarrow$ \[1 = \left( {1 - x} \right)\left( {1 + x + {x^2} + {x^3} + ...} \right.\]
So, \[\dfrac{1}{{1 - x}} = 1 + x + {x^2} + {x^3} + ... = \sum\limits_{k = 0}^\infty {{x^k}} \]
Note: In the given question, we had to find the Maclaurin series for a polynomial. We did that by keeping the constant \[1\] on one side of equality and the rest of the terms of the polynomial on the other. Then we started writing the multiplier for the polynomial so that it became equal to \[1\]. Then we generalized the series so as to get the required Maclaurin series for the given polynomial. So, it is really important that we know the formulae and where, when and how to use them so that we can get the correct result.
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