what's up here?

Dr.Peterson said:
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.


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.
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I am using the TI Nspire CAS. Tomorrow I will post some photos I took of what I did with the calculator.
 
JeffM said:
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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I don't use the calculator unless I know how to solve the thing manually. This problem was supposed to be solved manually and that is what I tried to do. I used the calculator merely to check my work. It is great for that. Sometimes the book tells us to use the calculator.

Re: Calculators: I wonder why universities don't give courses and even degrees and even higher degrees in the art of using calculators in math. The professors will know when and how to use the devices and all should go swimmingly. As things stand large multinats like Texas Instruments and HP put out marvelously ingenious and speed-of-light fast devices, the Nspire series and HP with the HP50 and the Prime and then take absolutely, or so it seems, no care to really educate users. The manuals they provide are all in pdf format and usually not link-indexed so that to find something you first have to look through ten pages of the table of contents, find the subject you want and then scroll down to the page to get the information (if it is even there). This is deplorable. Oh, for the days when you bought a device like the HP42 or 48 or 50 and received a fat paper manual to go with it. I didn't use calculators back then but I know the history as I am sure many of the contributors at this site do and much better than I.
 
JeffM said:
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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Yes. I see it. 1/3 has to be used throughout. I will go over this again tomorrow. Funny how such a little thing is such a big one and so easy to miss at least for a math practitioner who is not terribly experienced.
 
so, armed with the sharp sword of new knowledge I went back into the fray today and prevailed thus:
summ fin.PNG

Here are some of the images from my calculator that I used to check my answers while working on these problems:
summ2.PNG

summ3.PNG
 
In the 1st image you have two summations listed. I ask you to answer my questions below.
1) Do not look at the fraction results at all! Just looking at the summations. Can you tell me how much the two summations should differ by?
2) Can you please subtract the two fractions and post this result?

Are the two answers the same? Should they be? If not, then how can this be?
 
Jomo said:
In the 1st image you have two summations listed. I ask you to answer my questions below.
1) Do not look at the fraction results at all! Just looking at the summations. Can you tell me how much the two summations should differ by?
2) Can you please subtract the two fractions and post this result?

Are the two answers the same? Should they be? If not, then how can this be?
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they should differ by 1. I will do the calculations you suggest tomorrow and post my results.
 
Jomo said:
Thank you and yes, they should differ by 1. Given that you know that then what is the problem, if any, that you are having.
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well, the problem had to do with the way I was multiplying by 1/3 and dividing by -3. Somehow I was thinking these two figures could be convertible in this procedure. Not so.
 
Another complaint- at one point you have \(\displaystyle 3^{-k}\) and at another you have \(\displaystyle \frac{1^k}{3}\). Those are NOT the same!
 
allegansveritatem said:
well, the problem had to do with the way I was multiplying by 1/3 and dividing by -3. Somehow I was thinking these two figures could be convertible in this procedure. Not so.
Click to expand...
You have a signed number (a positive or negative number) and when you multiply it by 1/3 (which is positive) the sign of the result stays the same.
If you have a signed number and divide it by -3 (a negative number) then the sign of the result will change.
The facts above show that multiplying a number by 1/3 and dividing by -3 can't be the same.
 
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