# Question Video: Finding the Power Series for a Rational Function Mathematics • 12th Grade

Find a power series representation for the function π(π₯) = π₯β΄/(π₯ + 7).

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### Video Transcript

Find a power series representation for the function π of π₯ is equal to π₯ to the fourth power divided by π₯ plus seven.

The question is asking us to find a power series representation for our rational function π of π₯. To write our function π of π₯ as a power series, we want to show that π of π₯ is equal to the sum from π equals zero to β of π π multiplied by π₯ minus π all raised to the πth power for some sequence π π and some constant π. Since weβre asked to find the power series representation of a rational function, we can recall the following facts about geometric series.

The sum from π equals zero to β of π times π to the πth power is equal to π divided by one minus π if the absolute value of π is less than one. And this series diverges if the absolute value of π is greater than one. This gives us a method of turning a quotient into a power series. So, we want to write π₯ to the fourth power divided by π₯ plus seven in the form π divided by one minus π. Since we have a single π₯ term in our numerator, weβll just take π₯ to the fourth power outside of our fraction. We want to rewrite our denominator π₯ plus seven in the form one minus π. This means instead of seven, we want a constant one.

To achieve this, weβll take a factor of seven outside of our denominator. This gives us π₯ to the fourth power multiplied by one divided by seven times π₯ over seven plus one. We can then take our factor of one-seventh outside of our fraction. This gives us π₯ to the fourth power divided by seven multiplied by one divided by one plus π₯ over seven. And we can see one divided by one plus π₯ over seven is of the form π divided by one minus π. We have our initial value of π equal to one and our ratio of successive terms, π, equal to negative π₯ divided by seven.

So, by using our method to find the infinite sum of a geometric series, we have our function π of π₯ is equal to π₯ to the fourth power divided by seven multiplied by the sum from π equals zero to β of one times negative π₯ over seven all raised to the πth power. If the absolute value of negative π₯ over seven is less than one. We can simplify this. Multiplying by one does not change the value of our summand. And we can bring π₯ to the fourth power over seven inside of our sum. We can then also rewrite negative π₯ over seven as negative one times one over seven times π₯. This will then let us distribute the exponent over the parentheses.

Distributing the exponent over our parentheses gives us the sum from π equals zero to β of π₯ to the fourth power over seven multiplied by negative one to the πth power multiplied by one-seventh to the πth power multiplied by π₯ to the πth power. We have π₯ to the fourth power multiplied by π₯ to the πth power is π₯ to the power of π plus four. We can rewrite one-seventh all raised to the πth power as one divided by seven to the πth power. We then see that one-seventh multiplied by one over seven to the πth power is one over seven to the power of π plus one.

Rearranging our summand gives us that we found a power series representation for our function. π of π₯ is equal to π₯ to the fourth power divided by π₯ plus seven is given by. The sum from π equals zero to β of negative one to the πth power divided by seven to the power of π plus one multiplied by π₯ to the power of π plus four.