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

Find a power series representation for the function 𝑓(π‘₯) = π‘₯⁴/(π‘₯ + 7).

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Video Transcript

Find a power series representation for the function 𝑓 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 𝑓 of π‘₯. To write our function 𝑓 of π‘₯ as a power series, we want to show that 𝑓 of π‘₯ is equal to the sum from 𝑛 equals zero to ∞ of 𝑐 𝑛 multiplied by π‘₯ minus π‘Ž all raised to the 𝑛th power for some sequence 𝑐 𝑛 and some constant π‘Ž. Since we’re asked to find the power series representation of a rational function, we can recall the following facts about geometric series.

The sum from 𝑛 equals zero to ∞ of 𝑏 times π‘Ÿ to the 𝑛th power is equal to 𝑏 divided by one minus π‘Ÿ if the absolute value of π‘Ÿ is less than one. And this series diverges if the absolute value of π‘Ÿ 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 𝑏 divided by one minus π‘Ÿ. Since we have a single π‘₯ term in our numerator, we’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‘Ÿ. This means instead of seven, we want a constant one.

To achieve this, we’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 𝑏 divided by one minus π‘Ÿ. We have our initial value of 𝑏 equal to one and our ratio of successive terms, π‘Ÿ, 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 𝑓 of π‘₯ is equal to π‘₯ to the fourth power divided by seven multiplied by the sum from 𝑛 equals zero to ∞ of one times negative π‘₯ over seven all raised to the 𝑛th 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 𝑛 equals zero to ∞ of π‘₯ to the fourth power over seven multiplied by negative one to the 𝑛th power multiplied by one-seventh to the 𝑛th power multiplied by π‘₯ to the 𝑛th power. We have π‘₯ to the fourth power multiplied by π‘₯ to the 𝑛th power is π‘₯ to the power of 𝑛 plus four. We can rewrite one-seventh all raised to the 𝑛th power as one divided by seven to the 𝑛th power. We then see that one-seventh multiplied by one over seven to the 𝑛th power is one over seven to the power of 𝑛 plus one.

Rearranging our summand gives us that we found a power series representation for our function. 𝑓 of π‘₯ is equal to π‘₯ to the fourth power divided by π‘₯ plus seven is given by. The sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power divided by seven to the power of 𝑛 plus one multiplied by π‘₯ to the power of 𝑛 plus four.

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