what's up here?

A

allegansveritatem

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Here is the exercise: (I am being asked to find the sum):
summation.PNG

Here is what I did and what the calculator said about it:
summation1.PNG

I have worked this out in several ways. I have done the numerator and the denominator calculations separately and then divided them and still came up with the same answer. I tried changing the sign of the exponent and got a solution that was still not the one I got from the calculator. Where is the problem here?
 
Let’s go back to basics.

What is [MATH]3^{-k}[/MATH]?

[MATH]3^{-k} = (3^{-1})^k = \left ( \dfrac{1}{3} \right )^k.[/MATH]
What are you being asked to do? Simply add a small number of decreasing fractions. Your answer, which exceeds 300, does not pass the plausibility test.

[MATH]\dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{27} + \dfrac{1}{81} + \dfrac{1}{243} + \dfrac{1}{729} + \dfrac{1}{2187} =[/MATH]
[MATH]\dfrac{729 + 243 + 81 + 27 + 9 + 3 + 1}{2187} = \dfrac{1093}{2187}.[/MATH]
Now doing this sort of summation is tedious. There is a nice formula, but formulas are dangerous.

[MATH]r \ne 1 \implies \sum_{j=0}^n r^j = \dfrac{r^{(n+1)} - 1}{r - 1} \equiv \dfrac{1 - r^{(n+1)}}{1 - r}.[/MATH]
If we use that formula, we must, in the first place, use 1/3 rather than 3. I am not sure what you were thinking there.

[MATH]\dfrac{1 - \left ( \dfrac{1}{3} \right )^{(7+1)}}{1 - \dfrac{1}{3}} = \dfrac{\dfrac{6560}{6561}}{\dfrac{2}{3}} = \dfrac{3280}{2187}.[/MATH]
But that answer is WRONG.

The reason for the error is that the very nice formula does not directly apply to this problem because we are not starting from (1/3)^0.
We must adjust the results of the formula to make it fit the problem

[MATH]\dfrac{3280}{2187} - \left ( \dfrac{1}{3} \right )^0 = \dfrac{3280}{2187} - 1 = \dfrac{3280 - 2187}{2187} = \dfrac{1093}{2187}.\checkmark[/MATH]
 
JeffM said:
Let’s go back to basics.

What is [MATH]3^{-k}[/MATH]?

[MATH]3^{-k} = (3^{-1})^k = \left ( \dfrac{1}{3} \right )^k.[/MATH]
What are you being asked to do? Simply add a small number of decreasing fractions. Your answer, which exceeds 300, does not pass the plausibility test.

[MATH]\dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{27} + \dfrac{1}{81} + \dfrac{1}{243} + \dfrac{1}{729} + \dfrac{1}{2187} =[/MATH]
[MATH]\dfrac{729 + 243 + 81 + 27 + 9 + 3 + 1}{2187} = \dfrac{1093}{2187}.[/MATH]
Now doing this sort of summation is tedious. There is a nice formula, but formulas are dangerous.

[MATH]r \ne 1 \implies \sum_{j=0}^n r^j = \dfrac{r^{(n+1)} - 1}{r - 1} \equiv \dfrac{1 - r^{(n+1)}}{1 - r}.[/MATH]
If we use that formula, we must, in the first place, use 1/3 rather than 3. I am not sure what you were thinking there.

[MATH]\dfrac{1 - \left ( \dfrac{1}{3} \right )^{(7+1)}}{1 - \dfrac{1}{3}} = \dfrac{\dfrac{6560}{6561}}{\dfrac{2}{3}} = \dfrac{3280}{2187}.[/MATH]
But that answer is WRONG.

The reason for the error is that the very nice formula does not directly apply to this problem because we are not starting from (1/3)^0.
We must adjust the results of the formula to make it fit the problem

[MATH]\dfrac{3280}{2187} - \left ( \dfrac{1}{3} \right )^0 = \dfrac{3280}{2187} - 1 = \dfrac{3280 - 2187}{2187} = \dfrac{1093}{2187}.\checkmark[/MATH]
Click to expand...
but I thought k=1 under the sigma. K=0? That is what was specified in the exercise. I will have to think about what happens when 0 is the first operand (?) here. I see that when I put either of these expressions into the calculator I get the same thing back so...
 
JeffM said:
Let’s go back to basics.

What is [MATH]3^{-k}[/MATH]?

[MATH]3^{-k} = (3^{-1})^k = \left ( \dfrac{1}{3} \right )^k.[/MATH]
What are you being asked to do? Simply add a small number of decreasing fractions. Your answer, which exceeds 300, does not pass the plausibility test.

[MATH]\dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{27} + \dfrac{1}{81} + \dfrac{1}{243} + \dfrac{1}{729} + \dfrac{1}{2187} =[/MATH]
[MATH]\dfrac{729 + 243 + 81 + 27 + 9 + 3 + 1}{2187} = \dfrac{1093}{2187}.[/MATH]
Now doing this sort of summation is tedious. There is a nice formula, but formulas are dangerous.

[MATH]r \ne 1 \implies \sum_{j=0}^n r^j = \dfrac{r^{(n+1)} - 1}{r - 1} \equiv \dfrac{1 - r^{(n+1)}}{1 - r}.[/MATH]
If we use that formula, we must, in the first place, use 1/3 rather than 3. I am not sure what you were thinking there.

[MATH]\dfrac{1 - \left ( \dfrac{1}{3} \right )^{(7+1)}}{1 - \dfrac{1}{3}} = \dfrac{\dfrac{6560}{6561}}{\dfrac{2}{3}} = \dfrac{3280}{2187}.[/MATH]
But that answer is WRONG.

The reason for the error is that the very nice formula does not directly apply to this problem because we are not starting from (1/3)^0.
We must adjust the results of the formula to make it fit the problem

[MATH]\dfrac{3280}{2187} - \left ( \dfrac{1}{3} \right )^0 = \dfrac{3280}{2187} - 1 = \dfrac{3280 - 2187}{2187} = \dfrac{1093}{2187}.\checkmark[/MATH]
Click to expand...
as for the way I performed the calcu7lation, which topic is a fertile field for error, I did it thus 1 x 1-r^n over 3 times 1-r. I mean I multiplied numerator with numerator and denominator with denominator. How else?
 
the horrible truth just dawned on me: I multiplied by 1/3 instead of 1/3^n! Oh boy...does that little ^n make a big difference! I have to run now but later I will test this out. Thanks so much for your replies. Sometimes I get stuck in these little loops of folly and need help getting back out again.
 
allegansveritatem said:
the horrible truth just dawned on me: I multiplied by 1/3 instead of 1/3^n! Oh boy...does that little ^n make a big difference! I have to run now but later I will test this out. Thanks so much for your replies. Sometimes I get stuck in these little loops of folly and need help getting back out again.
Click to expand...
That is why it is super important to learn to DERIVE the formulas - like JeffM did. I have had to add up AP and GP at least 1000 times in my career. But every time I derive the equation. Then check my derivation with rote-memorized equations (or fancy calculators).
 
I think that you used your formula in an invalid way. I believe that you never verified that r = 3-1 was less than 1. Your formula only works if |r|<1. I believe that you never verified this because you never wrote 1/3 for 3-1.

[math]\sum_{k=1}^nx_n = (\sum_{k=0}^nx_n) - x_0[/math]
 
I get the little loops of folly. Been there many times myself.

But there are several morals to this story.

(1) Unless you ask someone who has the same type of calculator you are using, it is very difficult to answer questions that involve “why did my calculator give me this answer.”

(2) When someone goes to the trouble to write out a boring arithmetic problem validating your calculator’s answer, the calculator is almost certainly doing what the manual says will happen upon pressing some sequence of keys. So if your calculator starts at k = 1, that is fine. If the formula that you have memorized starts with k = 1, that is fine so long as you remember the formula and when it applies. But math is not so limited. There can be a formula that starts at k = 0. What your calculator does is fixed in electronics; those electronics do not limit math.

(3) Formulas are dangerous. I do not burden my memory with lots of formulas. I have memorized a few that are easy to remember and that I fully understand and can prove. And I disclosed the general formula that I used. It is easy to remember.

[MATH]\sum_{k=0}^n r^k = \dfrac{1 - r^{(n+1)}}{(1 - r)} = 1 + \sum_{k=1}^n r^k.[/MATH]
[MATH]\therefore \sum_{k=1}^n = \dfrac{1 - r^{(n+1)}}{1 - r} - 1 = \dfrac{r - r^{(n + 1)}}{1 - r} = \dfrac{r(1 - r^n)}{1 - r} = r * \dfrac{1 - r^n}{1 - r}.[/MATH]
So you remembered the formula correctly and are mis-remembering it now. (If you do not fully understand why a formula is what it is, you are apt to mis-remember it.)

But your formula is for positive powers, such as 1 to 7, not negative powers, such as -1 to -7. To make your powers positive, your r must be (1/3) rather than 3.

[MATH]\dfrac{1}{3} * \dfrac{1 - \left ( \dfrac{1}{3} \right )^7}{1 - \dfrac{1}{3}} = \dfrac{1}{3} * \dfrac{\dfrac{2186}{2187}}{\dfrac{2}{3}} = \dfrac{1093}{2187}.[/MATH]
 
Last edited:
@Jomo

I think you need to spend (1/3) seconds in the corner. The formulas work unless r = 1. The series does not have a limit unless

[MATH]|\ r \ | < 1.[/MATH]
I’m going to tell Subhotosh.
 
Last edited:
JeffM said:
@Jomo

I think you need to spend (1/3) seconds in the corner. The formulas work unless r = 1. The series does not have a limit unless

[MATH]|\ r \ | \le 1.[/MATH]
I’m going to tell Subhotosh.
Click to expand...
I thought we were friends. Oh well.
 
JeffM said:
@Jomo

I think you need to spend (1/3) seconds in the corner. The formulas work unless r = 1. The series does not have a limit unless

[MATH]|\ r \ | < 1.[/MATH]
I’m going to tell Subhotosh.
Click to expand...
Did you say that the series does not have a limit unless |r| < 1? What is the limit when r=1? How about r>1?
 
JeffM said:
@Jomo

I think you need to spend (1/3) seconds in the corner. The formulas work unless r = 1. The series does not have a limit unless

[MATH]|\ r \ | \leq 1.[/MATH]
I’m going to tell Subhotosh.
Click to expand...
NO, no, no, I already caught that error and reposted it. But I will be nice and not tell Subhotosh on you.
 
allegansveritatem said:
I have used the seq() command in my calculator with the formula 3^-n, summed up the results and they match what the calculator gave me for the sigma.
Click to expand...
What does your calculator have to do with anything? I asked for the first few terms to make sure you understand what's going on.
 
JeffM said:
I get the little loops of folly. Been there many times myself.

But there are several morals to this story.

(1) Unless you ask someone who has the same type of calculator you are using, it is very difficult to answer questions that involve “why did my calculator give me this answer.”

(2) When someone goes to the trouble to write out a boring arithmetic problem validating your calculator’s answer, the calculator is almost certainly doing what the manual says will happen upon pressing some sequence of keys. So if your calculator starts at k = 1, that is fine. If the formula that you have memorized starts with k = 1, that is fine so long as you remember the formula and when it applies. But math is not so limited. There can be a formula that starts at k = 0. What your calculator does is fixed in electronics; those electronics do not limit math.

(3) Formulas are dangerous. I do not burden my memory with lots of formulas. I have memorized a few that are easy to remember and that I fully understand and can prove. And I disclosed the general formula that I used. It is easy to remember.

[MATH]\sum_{k=0}^n r^k = \dfrac{1 - r^{(n+1)}}{(1 - r)} = 1 + \sum_{k=1}^n r^k.[/MATH]
[MATH]\therefore \sum_{k=1}^n = \dfrac{1 - r^{(n+1)}}{1 - r} - 1 = \dfrac{r - r^{(n + 1)}}{1 - r} = \dfrac{r(1 - r^n)}{1 - r} = r * \dfrac{1 - r^n}{1 - r}.[/MATH]
So you remembered the formula correctly and are mis-remembering it now. (If you do not fully understand why a formula is what it is, you are apt to mis-remember it.)

But your formula is for positive powers, such as 1 to 7, not negative powers, such as -1 to -7. To make your powers positive, your r must be (1/3) rather than 3.

[MATH]\dfrac{1}{3} * \dfrac{1 - \left ( \dfrac{1}{3} \right )^7}{1 - \dfrac{1}{3}} = \dfrac{1}{3} * \dfrac{\dfrac{2186}{2187}}{\dfrac{2}{3}} = \dfrac{1093}{2187}.[/MATH]
Click to expand...
I went back to work on this today and got tangled up again. I found one thing, however, that when I used 1 as the first term in the series then things worked out when I worked out the formula. So, is the calculator somehow doing that, I mean starting the series with 1? I know when I put the formula 1*3^-n into the seq() command I get 1/3 for first term. Also, you are right, I have to be careful that I understand the formulas I use.
 
Did you read post #9 carefully.

I will try to be more clear. [math]\sum _{i=1}^\infty x_i[/math] = x1+ x2 + x3 + ... = [x0 +x1+ x2 + x3 + ...] - x0 = [math](\sum_{i=0}^\infty x_i) -x_0[/math].

Basically I added the x0 term into the summation because I have a formula that works for this setup but then I needed to subtract that term, x0, in the end.

What I did can be used for geometric progression problems as well as other type of summation problems.
 
allegansveritatem said:
Here is the exercise: (I am being asked to find the sum):
View attachment 24123

Here is what I did and what the calculator said about it:
View attachment 24124

I have worked this out in several ways. I have done the numerator and the denominator calculations separately and then divided them and still came up with the same answer. I tried changing the sign of the exponent and got a solution that was still not the one I got from the calculator. Where is the problem here?
Click to expand...
You used the right formula, [MATH]a\frac{1-r^n}{1-r}[/MATH], and the right [MATH]a[/MATH]; you just forgot to use [MATH]3^{-1}[/MATH] for [MATH]r[/MATH].

The calculator is correct.

But if you don't show how you used your calculator (and what calculator it is), there's no way anyone could tell you if you used it wrong. The same is true of how you did the formula calculation.

allegansveritatem said:
So, is the calculator somehow doing that, I mean starting the series with 1? I know when I put the formula 1*3^-n into the seq() command I get 1/3 for first term. Also, you are right, I have to be careful that I understand the formulas I use.
Click to expand...
I have no idea what the seq() command is, or what you entered, and certainly wouldn't guess that's what you used to get the sum.
 
To summarize my posts, the formulas that we are discussing work for sums where the powers are successive non-negative integers. Your fundamental problem was not working with powers of 1/3 rather than three to turn the exponents into positive integers.

The formula that I have memorized deals with a series where the initial exponent is zero. The one you seem to be familiar with starts deals with a series where the initial exponent is one. It is easy to derive one formula from the other; there is no inconsistency between tham. I have found the formula summarizing a series starting with an exponent of zero is easier to remember, but that is mere personal preference. In this case, the formula summarizing a series starting from an exponent of one is directly applicable to the given problem. There was nothing wrong with the formula you elected to use, but you needed to work consistently with (1/3) rather sometimes 3 and sometimes (1/3). By the way, the usual derivation of your formula is very elegant, but, in my opinion, it gives little or no insight into power series.

Finally, working with calculators avoids a lot of tedious work. But it comes at a double cost. We can never be sure that we entered stuff in the device correctly, which is why reasonable checks are crucial. Moreover, it tempts us to just start banging away on the calculator without giving a problem much thought. In the very first response, lev suggested working the problem out partially by hand, which I then did in full. And that resulted in confirming the calculator’s result. From that point on, the calculator issue was irrelevant. All that was relevant was what formula you were using and were you using it correctly.
 
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