# Sigma notation and calculator help(casio fx9860g; ti-30xpro)

1. Sep 11, 2016

### late347

1. The problem statement, all variables and given/known data
A) use sigma notation and calculate sum as far as you wish (set your own endpoint)

(1/2) + (1/4) + (1/8) + (1/16)...

B) this is a personal problem because I want to know how to calculate the sums with e.g. sigma notation, with a calculator.

Never having done this before, I have two calculators available. An older model graphing calculator casio fx-9860G, and a calculator Ti-30X Pro multiview. I was wondering if anybody knew how to input the conditions of the sigma, and get the correct type of calculation into the calculator.

I do not posess the manual book any longer, for casio fx-9860G but that particular calculator has some nostalgia value to me, and I know how to use it in many other situations.

2. Relevant equations

$\sum\limits_{k=1}^n a_k= a_1 +a_2+a_3....+a_n$

3. The attempt at a solution

A)

I was confused about the general form formula for sigma.

How can I make sure, that the "pattern variable" for the purpose of creating the n-th term of the series, is the correct variable from sigma notation itself?

in other words if the sigma starts such as $\sum\limits_{k=1}^3 2^k$
does that equal to:
$2^1 + 2^2 + 2^3$

If instead I wrote sigma such as $\sum\limits_{k=1}^6 2^(n-1)$
does that equal to:
$2^(6-1=5)+ 2^(5) + 2^5 +2^(5) +2^(5) +2^(5)$
because the series is limited from the first term, until the sixth term, where n=6 initially, but gets reduced by one when calculating? If the code looks bad it was supposed to be such that (n-1) = exponent

1. the top number stays as 1 in each term of the series (does it matter if you call sigma sum, as series, is there any difference between a sum and a series like in this example?). If I remember correctly, sequences are simply numbers which follow some pattern, but the numbers are simply divided by a comma, and no mathematical operations as such are performed after creation of the terms of the sequence.

2. the bottom number changes I think in the form of 2x, where x is the index number of the term, which we are calculating. starting index being 1 for the first term.

such as the first term = $\frac{1}{2^1}$
2nd. term= $\frac{1}{2^2}$
3rd term = $\frac{1}{2^3}$

This leads me to conclude thatthe sigma notation can now be created, because we know the pattern for creation of new terms for the series.

$\sum\limits_{i=1}^n \frac{1}{2^i} = ~~ \frac{1}{2^1}~~+ ~~ \frac{1}{2^2}~~+~~ \frac{1}{2^3}~~...$

the sum for the first 5 terms calculated manually = 0.5 + 0.25 +0.125 +(1/16) +(1//32)=31/32

Last edited: Sep 11, 2016
2. Sep 11, 2016

### Math_QED

A) We have the sequence:

$\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \dots$

They ask you to write this sum in sigma notation and you can choose the endpoint yourself, then calculate the sum. So, you can be smart and choose the endpoint very small, so you don't need to calculate a lot. You calculated the sum for 5 terms, but you could also do the same for 1 term.

We then have: $\sum\limits_{i=1}^1 \frac{1}{2^i} = 1/2$. What you did was:
$\sum\limits_{i=1}^5 \frac{1}{2^i} = \frac{43}{8}$, which is NOT correct. (Recalculate that sum!)

Yes.

For the code, write the exponent between {} instead of () (I fixed your code in the quote).
No. You should specify that n = 6. Otherwise:

$\sum\limits_{k=1}^6 3^{n-1} = 3^{n-1} + 3^{n-1} + 3^{n-1} + 3^{n-1} + 3^{n-1} + 3^{n-1} = 6*3^{n-1} = 2*3*3^{n-1} = 2*3^n$. There is nothing that says that n is the number on top of the $\Sigma$ sign.

For the definition of a series: https://en.wikipedia.org/wiki/Series_(mathematics)

Last edited: Sep 11, 2016
3. Sep 11, 2016

### late347

ok i had a brain fart and the answer is 31/32 for the sum of 1-->5 terms

well ok I get now that n does not need to be the topcondition of the sigma.

however, if n were the top condition of the sigma, would that essentially be correct interpretation.

$\sum\limits_{k=1}^5 2^{n-1}= 2^4+2^4+2^4+2^4+2^4$

my textbook used the sigma notation such that
n= index of the upper bound term of the sum, included
k= index of the first term of the sum, included

Last edited: Sep 11, 2016
4. Sep 11, 2016

### late347

my textbook shows
$\sum\limits_{k=m}^{n}a_k=a_m +a_{m+1}+...a_{n-1}+a_{n}$
where a_n = final term
and where a_m = the first term (ostensibly???)

with term of the sum being for example a_k = 2^k
Normally the sum would be something like
$\sum\limits_{k=1}^{5}2^k= 2^1+2^2+2^3+2^4+2^5$
Now if I wanted to calculate sum from the third term, until the fifth term (3rd + 4th + 5th)
If I wanted to calculate the sum from the 3rd term until the 5th term, could I do that with sigma notation (skip counting the 1st and the 2nd terms of the sum)

5. Sep 11, 2016

### Math_QED

Yes it is 31/32.

No, you have to specify either way that n = 5. That they call it n is just arbitrary, it could be any letter that you want it to be. If you would state it this way, then it's much clearer: $\sum\limits_{k=1}^n 2^{n-1}$. Now it's obvious that if n = 5, then $\sum\limits_{k=1}^5 2^{5-1}= 2^4+2^4+2^4+2^4+2^4$. I hope this helps.

6. Sep 11, 2016

### late347

Well since you talked about being very specific with the n variable as e.g. the upperbound of the sum I could easily write on paper such that
$\sum\limits_{k=1}^{n=4}3^n=3^4+3^4+3^4+3^4$
Latex code can get a bit annoying at times, of course. Whether that above, is standard convention or not.

I think it would be accurate but a little bit silly example admittedly. If you can put k=1 on the bottom of the sigma why cant you put n=(whatever natural number) on the top?

but what about the next question.

$\sum\limits_{k=3}^{5}2^k=$ $2^3+2^4+2^5$ is this so???

7. Sep 11, 2016

### Math_QED

To avoid confusion, just write:
$n = 4 \Rightarrow \sum\limits_{k=1}^{4}3^n=3^4+3^4+3^4+3^4$

You can. As long as n>k or n = k.

Yes.

8. Sep 11, 2016

### late347

I figured out on my own how to use sigma sums on my newer texas instruments Ti -30x pro calculator.

However I was unable sadly to understand how to calculate sums with casio fx-9860g

9. Sep 11, 2016