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02202011  #1 
Join Date: Nov 2010
Posts: 36

Peter's GP Log question
Here is something I found by poking around with Peter's Series.
Question is why does it work (sort of!). ie 1/3  1/9 + 1/27... Suppose I give you a term from it say 1/6561. Now we know the common ratio could be calculated as follows: term 2 / term 1 = 1/9 / 1/3 = 1/3 so common ratio = 1/3 we also know the first term = 1/3 Which means we could use the formula for finding a given term from the series ar^(n1) Anywho here is the crux dear reader.... I could write 1/6561 = 1/3 * (1/3)^(n1) (Just to be clear in words the RHS is a third times minus a third raised to the power of n minus 1.) If I solve for n using logs I get 1 * log 1/6561 = log 1/3 + 1(n1)log 1/3 log 1/6561 = log 1/3 + (1n)log 1/3 log 1/6561 = log 1/3 + log 1/3 nlog 1/3 (log 1/6561  2log 1/3) / log 1/3 = n = 10 (which is the wrong answer) However, If I do the following I get an answer which is closer to the truth: So again I write 1/6561 = 1/3 * (1/3)^(n1) this time I multiply through by 1 and change the power around as follows: 1/6561 = 1/3 * (1/3)^(1n) log 1/6561 = 1.log 1/3 + (1n)log 1/3 log 1/6561 = log 1/3 +log 1/3 nlog 1/3 log 1/6561 = nlog 1/3 log 1/6561 / log 1/3 = n n = 8 Now we can't have a negative term. So I could say we want the absolute value of n and therefore we get n = 8. Ie 1/6561 is the 8th term in the series. Which is correct. The trick seems to work but why does is work? (at least sort of!). But why? Is there a better way of doing it? 
02222011  #2  
Join Date: Dec 2008
Posts: 249

Quote:
In your attempt to solve the problem using logs, your log step is incorrect. (1/6561) = (1/3)(1/3)^(n  1) Now you have to take the logarithm of both sides which gives: log(1/6561) = log[(1/3)(1/3)^(n  1)] NOT 1 * log 1/6561 = log 1/3 + 1(n1)log 1/3 which is what you have. log(1/6561) is not equal to 1*log(1/6561) As you probably know taking the log of a negative number is not defined, so you cannot use that method. 

02232011  #3  
Join Date: Nov 2010
Posts: 36

GP and Logs
Quote:
I've had a rethink and come up with the following: If we ignore the negatives in all the variables to the equation as follows it seems to do the trick. Example 1 consider Peter's GP. 1/3  1/9 + 1/27... we know the common ratio is 1/3 and the first term is 1/3. Now, suppose we are given the following term from the series 1/1594323 and were asked to find the number of the term. term = a(r)^n1 and solve for n (the number of the term remember!) so in our question we have  First term = 1/3 (+ve no problem)  Common ratio = 1/3 (so take absolute value = 1/3 for our purposes)  Finally the term that we are trying to find the number of 1/1594323 and again take the absolute value we get + 1/1594323. Slot them into the eqn above and we get: 1/1594323 = 1/3 * (1/3)^(n1) log 1/1594323 = log 1/3 +(n1)log 1/3 log 1/1594323 = log 1/3 + nlog 1/3 log 1/3 so n = (log 1/1594323  log 1/3 + log 1/3) / log 1/3 n = log 1/1594323 / log 1/3 n = 13 so 1/1594232 is the 13th term in the series. Example 2 Consider the following GP 1/3, 1/9, 1/27, 1/81, 1/243, 1/729... we can calculate the number of the term for 1/243. we know already it is the 5th term but can we find it with logs? First term = 1/3 becomes 1/3 Common ratio = 1/3 becomes 1/3 Term to find number of = 1/243 becomes 1/243 1/243 = 1/3 * (1/3)^(n1) log 1/243 = log 1/3 + (n1)log 1/3 Log 1/243 = log 1/3 + nlog 1/3 = log 1/3 log 1/243 = nlog 1/3 n = log 1/243 / log 1/3 n = 5 So 1/243 is the 5th term in the series. Example 3 Consider the following series: 7/9, 7/15, 7/25, 21/125... From the information above and the following term from the series 567/15625 determine the number of the term within the series. Common ratio = 3/5 becomes 3/5 first term = 7/9 becomes 7/9 term we need number of 567/15625 becomes 567/15625 So we have 567/15625 = 7/9 * (3/5)^(n1) log 567/15625 = log 7/9 + (n1)log 3/5 log 567/15625 = log 7/9 + nlog 3/5  log 3/5 n = (log 567/15625  log 7/9 + log 3/5) / log 3/5 n = 7 So 567/15625 is the seventh term in the series. check: 7/9, 7/15, 7/25, 21/125, 63/625, 189/3125, 567/15625 LIsa 

02232011  #4  
Join Date: Dec 2008
Posts: 249

Can you look at example 3 again? I think there is something wrong with it. It doesn't look like a geometric series.


02232011  #5  
Join Date: Nov 2010
Posts: 36

Quote:
GP from example 3 is derived as follows: Common ratio = 7/15 / 7/9 = 3/5 Hence we have the following terms: 1st term = 7/9 2nd term = 7/9 * 3/5 = 7/15 3rd term = 7/15 * 3/5 = 7/25 4th term = 7/25 * 3/5 = 21/125 5th term = 21/125 * 3/5 = 63/625 6th term = 63/625 * 3/5 = 189/3125 7th term = 189/3125 * 3/5 = 567/15625 Hence 7/9, 7/15, 7/25, 21/125, 63/625, 189/3125, 567/15625... Hope this helps! 

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