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

# Question Video: Finding the Second Derivative Using Implicit Differentiation of Trigonometric Equations Mathematics • Higher Education

Multiply the series 1/(1 + π₯) = β_(π = 0)^(β) (β1)^(π) π₯^(π) 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 π equals zero to β of negative one to the πth power π₯ to the πth 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, π of π₯ is equal to the sum over π of π π π₯ to the πth power and π of π₯ is equal to the sum over π of π π π₯ to the πth power, which both converge for this value of π₯. Then we can calculate the product of these two power series, π of π₯ multiplied by π of π₯, as the sum over π of all of our π₯ to the πth power terms with the coefficient the sum over π π π multiplied by π π minus π. 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, π of π₯ and π of π₯, to be equal to one divided by one plus π₯.

So we have both π of π₯ and π of π₯ are equal to the sum of one over one plus π₯ which are equal to the sum from π equals zero to β of negative one to the πth power π₯ to the πth power. So for the value of π₯ where our power series for π of π₯ converges, we can use our formula to find an expression for π of π₯ squared. The question tells us that one divided by one plus π₯ is equal to the sum from π equals zero to β of negative one to the πth power π₯ to the πth power. This gives us that π π is negative one raised to the power of π. And π π minus π is negative one raised to the power of π minus π.

Therefore, we can calculate π π multiplied by π π minus π as negative one to the power of π multiplied by negative one to the power of π minus π, which is just negative one to the πth power. So we now have our coefficient of π₯ to the power of π is the sum from π equals zero to π of negative one to the πth power. And we can see thatβs our summand, negative one to the πth power, is independent of π. So, in fact, our sum is negative one to the πth power added to itself π plus one times. This is because π goes from zero to π. And if weβre adding negative one to the πth power to itself π plus one times, this is the same as saying π plus one multiplied by negative one to the πth power.

Therefore, weβ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 π equals zero to β of negative one to the πth power multiplied by π plus one multiplied by π₯ to the πth power.

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