Video Transcript
Multiply the series one divided by
one plus 𝑥 is equal to the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th
power 𝑥 to the 𝑛th power by itself to construct a series for one divided by one
plus 𝑥 squared. Write the answer in sigma
notation.
We recall if we have two power
series, 𝑓 of 𝑥 is equal to the sum over 𝑛 of 𝑎 𝑛 𝑥 to the 𝑛th power and 𝑔 of
𝑥 is equal to the sum over 𝑛 of 𝑏 𝑛 𝑥 to the 𝑛th power, which both converge
for this value of 𝑥. Then we can calculate the product
of these two power series, 𝑓 of 𝑥 multiplied by 𝑔 of 𝑥, as the sum over 𝑛 of
all of our 𝑥 to the 𝑛th power terms with the coefficient the sum over 𝑗 𝑎 𝑗
multiplied by 𝑏 𝑛 minus 𝑗. In our case, we want to multiply
one divided by one plus 𝑥 by itself to get the power series for one divided by one
plus 𝑥 squared. So we set both of our functions, 𝑓
of 𝑥 and 𝑔 of 𝑥, to be equal to one divided by one plus 𝑥.
So we have both 𝑓 of 𝑥 and 𝑔 of
𝑥 are equal to the sum of one over one plus 𝑥 which are equal to the sum from 𝑛
equals zero to ∞ of negative one to the 𝑛th power 𝑥 to the 𝑛th power. So for the value of 𝑥 where our
power series for 𝑓 of 𝑥 converges, we can use our formula to find an expression
for 𝑓 of 𝑥 squared. The question tells us that one
divided by one plus 𝑥 is equal to the sum from 𝑛 equals zero to ∞ of negative one
to the 𝑛th power 𝑥 to the 𝑛th power. This gives us that 𝑎 𝑗 is
negative one raised to the power of 𝑗. And 𝑏 𝑛 minus 𝑗 is negative one
raised to the power of 𝑛 minus 𝑗.
Therefore, we can calculate 𝑎 𝑗
multiplied by 𝑏 𝑛 minus 𝑗 as negative one to the power of 𝑗 multiplied by
negative one to the power of 𝑛 minus 𝑗, which is just negative one to the 𝑛th
power. So we now have our coefficient of
𝑥 to the power of 𝑛 is the sum from 𝑗 equals zero to 𝑛 of negative one to the
𝑛th power. And we can see that’s our summand,
negative one to the 𝑛th power, is independent of 𝑗. So, in fact, our sum is negative
one to the 𝑛th power added to itself 𝑛 plus one times. This is because 𝑗 goes from zero
to 𝑛. And if we’re adding negative one to
the 𝑛th power to itself 𝑛 plus one times, this is the same as saying 𝑛 plus one
multiplied by negative one to the 𝑛th power.
Therefore, we’ve shown for the
values of 𝑥 where a power series for one divided by one plus 𝑥 converges. We can multiply the power series of
one divided by one plus 𝑥 by itself to get that one divided by one plus 𝑥 squared
is equal to the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power
multiplied by 𝑛 plus one multiplied by 𝑥 to the 𝑛th power.
