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### Course: Integral Calculus>Unit 5

Lesson 14: Function as a geometric series

# Function as a geometric series

Expressions of the form a/(1-r) represent the infinite sum of a geometric series whose initial term is a and constant ratio is r, which is written as Σa(r)ⁿ. Since geometric series are a class of power series, we obtained the power series representation of a/(1-r) very quickly.

## Want to join the conversation?

• At , the sum of geometric sequence is a/(1-r), but this is only true for |r|<1, isn't it?
Applying this to f(x) seems like a bit of a stretch to me.
• Yes, you are correct. This series converges when |-x³| < 1 ⇔ |x| < 1.
• How can we say that Geometric series are Maclaurin series? or Maclaurin series are Power series? I think there's an information gap there. Can you explain to me?
• I don't like asking these type of questions but What is the point of this ?
what the point of representing a nice rational function with a series that only works when x is between (-1,1)
is this trick used in some proofs or when solving some questions if so can someone give me an example

second question is what if we wanted to have a taylor series and expand it at x=3 for example what would I do ?
• How is this a Maclaurin series? Doesn't the Maclaurin series have a factorial denominator in every term?
• Good question. If you take the first, second and third derivative of f(u), you can then evaluate f f' f'' f''' at u=0. That will give you the first four coefficients of the Maclaurin Series. Next, put each of those coefficients to their corresponding terms of the M. series using u as your variable - corresponding terms meaning u^0/0! + u^1/1! + u^2/2! + ... . You can then substitute every u back to x^3. Then simplify. The process will take only a few minutes - using u in place of x^3 - and you will quickly see that it reduces to the same equation as above, in Sal's video.
• How do you get rid of that autoplay button at the end..it doesn't let me rewind the video..so annoying..
• You can push the counter clockwise button in the down left corner to rewatch the video.
• at around , how do we know a Maclaurin series is a power series?
• Well, think about how a power series is defined. If I have a power series centred at c, it'd be the sum from n = 0 (or 1) to infinity of (a_n)(x-c)^n, where a_n is a real number.

See that this is exactly how a Maclaurin series looks like (just that the coefficients themselves need to be found out using a different formula). So, a Maclaurin (and in general, a Taylor) series is a kind of power series.
• When Sal makes this connection with the geometric series formula, how can we be sure it will be the Maclaurin Series? Is there only one unique Maclaurin Series for every function?
• Yes, there is only one unique Maclaurin series for every function. Maclaurin series are always constructed around the function where x=0. To check that this is the Maclaurin series for the function, plug x=0 into any partial sum of the Maclaurin expansion, and you will find that it is equal to the exact function.
(1 vote)
• Why is the power series not over n!
Are geometric series a special circumstance that look different from a normal Maclaurin series?
• The serie is p(x)=6 -6x^3 +6x^6......
the funtion is f(x)= 6/(1+x^3)
if x=1 f(1)=3 and p(1)=0 or 6
why f and p not equal in a infinite maclaurin's serie the error is zero?
• I don't think what you did was allowed I mean isn't this an oscillating sequence that diverges.
I think that you are not allowed to make x=1 in p(x) because we have a constriction that the ratio in a geometric sequence must be less than 1 so we have |x^3| < 1 meaning that
-1<x<1
a lot of contradiction happen when you allow p(1) = 3
I mean notice that I can simply can add p(1) = 6-6+6-6... and it wouldn't change the seires
2p(1)=p(1)
implying that 2=1 which is absurd.
maybe someone who is an expert can shed some more light on this topic.
(1 vote)
• at around , f'(x)=-18x^2/(1+x^3)^2.

substitute x=0, we get f'(0)=0.

according to the formula of the Maclaurin series (f^n(0)*x^n/n!), we get the 2nd term of the Maclaurin series to be 0, which is different from -6x^3.

Can anyone point out where I did wrong? Thanks.
(1 vote)
• You're not wrong. The second term will be 0. However, if you go ahead and find the third derivative, you get -36. So, the third term becomes -36 (x)^(3)/3! which is -6x^3.

Essentially, see that each term of the power series Sal derived has only powers which are multiples of 3. So, even when finding the Maclaurin series with your traditional method, you'll only get a term every third time (You'll get a term on the 3rd, 6th, 9th... derivatives and every other derivative will be 0). So, just to get 5 terms like Sal did, you'll need to take 12 derivatives, which is definitely not something you'd want to do. Hence, he proposed this method.