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Course: Calculus, all content (2017 edition)>Unit 7

Lesson 15: Power series function representation

Power series of ln(1+x³)

We can represent ln(1+x³) with a power series by representing its derivative as a power series and then integrating that series. You have to admit this is pretty neat. Created by Sal Khan.

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• Why did we do the integration bit ? As in what was the aim of it ?
• It's not a well motivated video.

I believe expanded forms are useful for computational approximations where accuracy is controlled by however many terms you choose to include.

In the real world you'd start with ln(1+x^3) and try to find its expanded form (approximation) by taking the derivative, expanding, and then integrating -- you wouldn't start with a geometric series and integrate its closed form because you're just looking for something "neat" to do.
• At , how do we know to start the summation at n=1 and not n=0? would we have to change the summation if we started at n=0?
• You have to create the summation so that it represents the series you have. You can start the summation at an n equal to whatever you want, but you would have to adjust the body of the summation to keep things the same.

The first term in the series is , the second is -x⁶/2, so that is what guides you.

If you decided to start the summation at n=0, then the body of the summation would have to be: (-1)ⁿ(x³ⁿ⁺³)/(n+1)

If you decided to start the summation at n=-1, then the body of the summation would have to be: (-1)ⁿ⁺¹(x³ⁿ⁺⁶)/(n+2)

Sal decided to start the summation at n=1 since that made the body of the summation quite simple.
• This pretty mathmatical art but i did not get what this can be useful for .....
• in the very last formula Is it okay to put (-1)^(n-1) ? instead of (-1)^(n+1) ?
• That is correct. Those expressions will have the same value for any value of n. So if the situation requires it, you could make that substitution.
• @ I don't think anybody will realize it's the derivative of natrual log of something lol
• calc so fun
• What happens to the interval of convergence after we integrate the polynomial expression to find a series for log(1+x^3)? Are we still constrained by the same interval or does it not apply anymore, and if it doesn't, why is that?
• dont you have to check the limits of the interval of convergence?
• No, because you express the result of the left hand side integral in "x" terms,
(1 vote)
• when trying to find the value of the constant of integration 'C', how was the value of x assumed to be 0 ? .... although -1 < x < 1 is our interval of convergence, is x = 0 a part of the interval?... wouldn't x = 0 make (-x^3) = 0 ? and if ( - x ^3) = 0 , wouldn't our common ratio become zero? and wouldn't the initial series cease to be a geometric series ?
• Sal wanted to use a value for x that would knock out all the power terms with x in them in order to solve for the C. He had removed all the unspecified terms and combined them into C2 to set the next step up.
A common strategy is to choose zero whenever you want to knock out a variable, as long as that was within the restricted domain of -1 < x < 1, and he mentioned he was checking, even though he knew very well that 0 is in that domain.
You asked whether it was: the domain -1 < x < 1 includes all numbers between -1 and 1. If we are counting by integers, the only integer in that domain is zero. If we are counting by smaller divisions than integers, we could use -.99999 -.889653 -.1111111 -.000002 0 0.000008, 0.55556, .999999 and an infinite number of other decimals or we could use
-½, -¼ , 0 , ¼, ½ or any combination just larger than -1 and just smaller than +1. The only really useful choice for this purpose, to avoid dealing with an infinite number of terms, is to use 0.
So
He started with his equation
ln (1 + x³) + C1 = C2 + x³ - ½x⁶ + ⅓x⁹ - ¼x¹²
He simplified it by subtracting C1 from both sides to get C3 or you can just call it C because you no longer have to differentiate between multiple unknown constants:
ln (1 + x³) = C + x³ - ½x⁶ + ⅓x⁹ - ¼x¹²
So, when x = 0, we have
ln (1 + 0) = C + 0 - 0 + 0 - 0
which is the same as
ln (1 ) = C + 0 - 0 + 0 - 0
Now, because the natural log of 1 is 0, this equation becomes
0 = C + 0 - 0 + 0 - 0
So C = 0

Your other question was wouldn't 0 make x³ = 0, the answer is
yes it would for that particular calculation, but the value of your common ratio does not change simply because you plugged in a value for one iteration of the calculations.

Hope that helps.