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### Course: APยฎ๏ธ Calculus BC (2017 edition)ย >ย Unit 12

Lesson 12: Power series function representation

# Geometric series as a function

Power series of the form Σk(x-a)โฟ (where k is constant) are a geometric series with initial term k and common ratio (x-a). Since we have an expression for the sum of a geometric series, we can rewrite such power series as a finite expression. Created by Sal Khan.

## Want to join the conversation?

• I'm a little bit confused... Does it mean that the infinite geometric series does not equal the function when x is outside the interval of convergence?
• If x is outside the interval of convergence, the function will still be a geometric series however it will not converge and the limit does not exist. This means that your geometric series does not add up to an obtainable value as you go towards infinity.
• I understand that this is the algebraic geometric series centered at 0. How can it be determined centered at another value, c, with the x-term becoming (x-c)? I couldn't find this explanation in this video series or by searching elsewhere. Does it exist in Khan Academy? Thank you.
• Here we see: โ 2*(-4x^2)^n, where the common ratio is (-4x^2). I would THINK that if you can break up your common ratio into the sum of an x-varying part minus a constant part, so it looks like (x-c), then that would show that the series is centered at the constant part.

Example: โ 2*[( 4b^2) - 5]^n. The common ratio is (4b^2)-5. If you made the substitution x = 4b^2. You could write f(x) = โ 2*(x-5)^n. I THINK you could reasonably say that this series is centered at c.
• Shouldn't Sal be testing the endpoints of the interval due to inconclusive ratio test at 1?
• No, because geometric series don't converge at the endpoints.
• Why didn't he test the inconclusive values to see if +/-1/2 should be included in the interval or not?
• Since we are dealing with a geometric series, the test to see whether or not the series converges is to see if the absolute value of the common ratio is less than 1. Therefore, per the test, x = +/- 1/2 diverges.
(1 vote)
• let me just understand something here ,i understand the concept of the interval of convergence and all that ,but is it still correct to say that a function is exactly equal to the expression (a/1-r) when r is within the interval of convergence ? what if we have that r is slightly less or more say (r+x) etc .but r+x was still within the radius of convergence .it would be incorrect to say that both [a/1-r ] and [a/1-(r+x)] are exactly equal to our function ,since that would imply they are equal to each other .So isn't the expression [a/1-a] just a really good approximation ?
(1 vote)
• The misunderstanding you have is that the function where r = r and the function where r = r + x are two different functions. Say the first one if f1, second one is f2. Then f1 = a/(1-r) and f2 = a/(1-(r+x)). The expression a/(1-r) is correct for each function since r is not the same for them.
• he is confusing me with absolute value thing. he should just take the number out straight up
(1 vote)
• Is it mathematically legal if the starting term 'a' is defined in terms of a variable (i.e. x^2)?
(1 vote)
• I would imagine so, since x^2 is actually a constant for a given x. But even if it weren't it would still be a thing that's possible, it just may or may not fit into the definition of a geometric series.
(1 vote)
• Can you express this function graphically as the sum of (โfrom n=-infinity to n=-1/2 of f(x)) + 2/(1+4x^2)+ (โfrom n=1/2 to n=infinity of f(x))?
(1 vote)
• What does ฮฃ (sigma) mean?
(I know it's Greek).
`` โ                 โ โ 2*(-4x^2)^n  =  โ 2*(-4x^2)^(n-1)n=0               n=1``