Video: Finding the Second Derivative Using Implicit Differentiation of Trigonometric Equations

Multiply the series 1/(1 + π‘₯) = βˆ‘_(𝑛 = 0)^(∞) (βˆ’1)^(𝑛) π‘₯^(𝑛) by itself to construct a series for 1/(1 + π‘₯)Β². Write the answer in sigma notation.

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Video Transcript

Multiply the series one divided by one plus π‘₯ is equal to the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power π‘₯ to the 𝑛th power by itself to construct a series for one divided by one plus π‘₯ squared. Write the answer in sigma notation.

We recall if we have two power series, 𝑓 of π‘₯ is equal to the sum over 𝑛 of π‘Ž 𝑛 π‘₯ to the 𝑛th power and 𝑔 of π‘₯ is equal to the sum over 𝑛 of 𝑏 𝑛 π‘₯ to the 𝑛th power, which both converge for this value of π‘₯. Then we can calculate the product of these two power series, 𝑓 of π‘₯ multiplied by 𝑔 of π‘₯, as the sum over 𝑛 of all of our π‘₯ to the 𝑛th power terms with the coefficient the sum over 𝑗 π‘Ž 𝑗 multiplied by 𝑏 𝑛 minus 𝑗. In our case, we want to multiply one divided by one plus π‘₯ by itself to get the power series for one divided by one plus π‘₯ squared. So we set both of our functions, 𝑓 of π‘₯ and 𝑔 of π‘₯, to be equal to one divided by one plus π‘₯.

So we have both 𝑓 of π‘₯ and 𝑔 of π‘₯ are equal to the sum of one over one plus π‘₯ which are equal to the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power π‘₯ to the 𝑛th power. So for the value of π‘₯ where our power series for 𝑓 of π‘₯ converges, we can use our formula to find an expression for 𝑓 of π‘₯ squared. The question tells us that one divided by one plus π‘₯ is equal to the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power π‘₯ to the 𝑛th power. This gives us that π‘Ž 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 π‘Ž 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 π‘₯ 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 π‘₯ where a power series for one divided by one plus π‘₯ converges. We can multiply the power series of one divided by one plus π‘₯ by itself to get that one divided by one plus π‘₯ 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 π‘₯ to the 𝑛th power.

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