A function that computes the sum of a geometric series.

A FUNCTION that computes the sum of a geometric series 1 + r + r^2 + r^3 + r^4 + ... + r^n, for a given r and N. THe input to the function must be 'r' and 'n'

Not sure what I am doing wrong, but I was trying to take baby steps and work it into a function but that didn't execute.

% create a vector with n elements all identical to r
 v = r*ones(1,n);
 % calculate [r r^2 r^3….r^n]
 v = cumprod(v);
 % sum and add one
 geoSum = 1 + sum(v);

[John D'Errico]

Please get used to using the "{} Code" button to format your code.

1. Paste in the code in the edit window.

2. Click on the "{} Code" button.

Your code will now be readable.

I've fixed your code this time.

Answer 1

[John D'Errico]

cumprod is a great idea. Well done for thinking of it.

You state that your function did not execute. This may be because it does not have a function header. All you have written is a script.

Have you confused n and N? The code you wrote assumes n.

So how did you call it? What error do you get? Show the full text of the error.

You've made a good start in what you did, so some variations that should work:

function S = geosum1(r,n)
  S = sum(cumprod([1,r*ones(1,n)]));

function S = geosum2(r,n)
  S = sum(cumprod([1,repmat(r,1,n)]));

It can also be done without cumprod.

function S = geosum3(r,n)
  S = sum(r.^(0:n));

Of course, you can break it into multiple lines. That makes the code more readable, and MATLAB does not charge extra if you use an extra line. For example:

 function S = geosum4(r,n)
   % sum of a geometric series, up to r^n, as
   %   1 + r + r^2 + ... + r^n
   % Note there will be n+1 terms in the series.

   % generate a vector to be then prodded together
   v = [1,r*ones(1,n)];

   % use cumprod, instead of using exponents
   % to compute each term as r^k
   p = cumprod(v);

   % sum the terms
   S = sum(p);

Comments are very important, as they help you to understand what you wrote, when you are forced to debug code written a year ago (or 30 years ago.) As well, too often once finds code written by a colleague, that you need to use. It can be crucial to be able to understand and follow their code if you will then use it and trust it.

Any of the above schemes will work. To verify that fact, we can even do this:

geosum3(sym('r'),3)
ans =
r^3 + r^2 + r + 1

[Jan]

+1. I like your style of teaching, which is useful for the readers.

And for completeness: There is a formula to determine the sum without accumulating the terms:

if r == 1
  S = n + 1;
else
  S = (1 - r ^ (n + 1)) / (1 - r)
end

[John D'Errico]

Yes. This is the classical solution for the sum of a geometric series, which is well worth understanding the derivation of, as the concept will appear more than once as a student learns mathematics.

Answer 2

[John BG]

Hi Patrick

clear all,clc

N=3
r=3

R=[1:1:r]
sr=0

for k=1:1:r
sr=sr+R(k).^[1:1:N]
end

result=sum(sr)

if you find this answer useful would you please be so kind to mark my answer as Accepted Answer?

To any other reader, please if you find this answer of any help,

please click on the thumbs-up vote link,

thanks in advance

John BG

<mailto:[e-mail] [e-mail]>

[John D'Errico]

Actually, John BG has made at least two mistakes here. I'll let him figure out what they are. And, please, don't just double down and insist that your answer is perfect. It is not even mathematically correct.

Then insisting that the asker must accept an incorrect answer is just wrong.

So read my answer, and think about what mistakes you made. Of course, if you need to read another answer to learn what you did wrong...

Anyway, Answers is NOT a competition. If you think it is, then you are wrong. Answers is about helping people who have a problem. Giving them help, with a correct answer, one that they can understand is what matters. As important is leaving behind an answer that may help others who have the same problem in the future.

[Roger Stafford]

Well said in your last paragraph, John D'Errico!! More than one contributor needs to hear that. I'll have to admit, there are times when I, too, get a bit over-competitive.

[John D'Errico]

I think it is easy to fall into the trap, something we ALL can do, without even noticing that we did. Maybe that is a flaw of compiling points. No system is perfect.

If our answer is a good one, correct, that teaches someone how to solve their problem, then that should be our only real reward. The "points" will arrive naturally then.

[Jan]

@Roger and John D'Errico: There is a cooperative kind of "competitions" also: when the motivation is to find the best answer for a question. I'm happy to see that the vast majority of submissions in this forum is constructive.

John D'Errico, during reading many of your answers --most of all when you decide wisely to assist at a homework question-- it is a pleasure to see and feel the motivation, that the question deserves a good answer - and not that the answerer deserves to get credit points. I like your style of teaching.

Roger, I've never read an over-competitive answer of you. You do not try to win on the costs of others and in all of the threads I remember, the asker was the winner.

It is disappointing when a member abuses the hospitality of this forum for getting personal kicks. The admins remove aggressive comments frequently and repeatedly, but currently I do not see a lastig effect. I appreciate your reactions to anti-social behavior, thank you very much. This supports the productive atmosphere of this forum and I really hope that I can spend my 2ct to aid this.

[John BG]

It is reasonable to consider that Mr Bradford wrote the question because

'r' and 'n' are/may be vectors not scalars

I have implemented the sum of N geometric series that in my opinion is the important point to solve in this question.

reading the question

 .. a geometric series 1 + r + r^2 + r^3 + r^4 + ... + r^n ..

r and n may be vectors, why couldn't them?

And what is the point of implementing a function when there is no such need for one? every one implicitly agrees on this point.

Mr d'Errico: Asking whether an answer may be considered accepted or not is not a mistake.

My answer is correct. Can you prove it isn't?

When time ago I found a basic bug in your variable length arithmetic toolbox I found it to be in the basics of politeness to let you know, as I did, and you appreciated my feedback in return, thanks.

Call it competition, or collaboration, or rivalry. Call it open market, or a battle field, or the Olympic games, or a Maths classroom:

Who wouldn't agree that only through competition improvement is achieved.

In this question, I have put effort to supply a working answer, with working MATLAB code to a question.

Jan Simon has also supplied working MATLAB code, including a refresher on the well known geometric series sum formula.

To Mr d'Errico: as useful and helpful and your comments are, you decided not to supply MATLAB working tested stand alone code to this question.

This forum is a really good thing, because people can ask anything they want, MATLAB related indeed, and people can answer what they consider to be valid working stand alone MATLAB answer, or simply add constructive helpful comments.

So, I would like to politely remark, just out of comment sense, please don't take it as a personal remark but as a broad line, that commenting things, in the context of people seeking working MATLAB code for MATLAB related questions, usually it is not as useful to question originators, as supplying MATLAB tested code in the answer.

Also glad to see that eventually every one is putting the needs of question originators ahead of rivalries: no more 'homework' tagging without consent of the person originating the question, or negative comments on whether a question is homework or shouldn't be asked in this forum, whenever there's reasonable doubt.

To all the forum gurus who read my answer to this question, thanks.

To Mr Bradford, perhaps you would like to consider marking the supplied answer of your choice that best fits as solution to your question.

Regards

John BG

<mailto:[e-mail] [e-mail]>

[Adam]

"My answer is correct. Can you prove it isn't?"

n = 3, r = 3 gives a problem of 1 + 3 + 3^2 + 3^3 = 1 + 3 + 9 + 27 = 40.

Your code returns 56 when I run it.

Or am I missing something?

As for other stuff, I think reputation and points systems are generally a bad idea, but in different forms they appear in most fora. Certainly people asking for their answer to be accepted or people to vote for it will always get people's backs up though, it is just rude and self-serving, but then again, a forum isn't a popularity contest either. Doing people's homework for them will always be popular with the person getting an easy homework answer!

[Jan]

 And what is the point of implementing a function when there is
 no such need for one? every one implicitly agrees on this point.

The wording of the question: "A FUNCTION that computes [...]" sounds exactly like a function is needed.

 To Mr d'Errico: [...] you decided not to supply MATLAB working
 tested stand alone code to this question.

John D'Errico has posted 4 working solutions and explained the differences exhaustively.

 My answer is correct. Can you prove it isn't?

Staying as near as possible to what was asked exactly:

n = 3; r = 3;
% Text of question:  1 + r + r^2 + r^3 + r^4 + ... + r^n
Result = 1 + 3 + 3^2 + 3^3;  % 40

Your code calculates 56. Assuming that 1:r and 1:N is meant does not seem natural, but even if you do this creating a scalar result is strange. This is not the "sum of a geometric series".

 Who wouldn't agree that only through competition improvement
 is achieved.

Me. I wouldn't agree. I trust in the productiveness of collaboration and mutual respect.

 no more 'homework' tagging without consent of the person
 originating the question

No, I do not agree. I answer homework questions in the forum exactly as if a student asks me the same question face to face. The reputation of this forum would suffer, if it becomes a platform for homework cheaters. Especially John D'Errico acts very wise with homework question and posts solution only, when the author shows effort and ask a specific question - exactly as a professor would react to a elaborate question of a student. The questions tagged as "homework" are "do-it-4-me" question which usually only cite the original assigment question. This is something complete different and the "let-me-do-it-4-you" answers are counter-productive. This is my opinion.

John D'Errico's answer gets clear, when we consider the other discussions of the last months, which have been repeatedly removed by the admins.

[John BG]

It's correct when n and r are vectors.

So you do not want n and r to be vectors, just scalars.

For a single series, not n series, just 1 series, then:

.

By all means, may I suggest you mark Jan Simon's answer as the accepted answer to your question?

regards

John BG

<mailto:[e-mail] [e-mail]>

[John D'Errico]

"It's correct when n and r are vectors."

No. It is not correct as you wrote it in your answer. Stop insisting that you could never possibly make a mistake, and think about what you wrote.

I find it interesting that you claim that r could be a vector. Yet your solution has completely invalid code when r is indeed a vector. Will the code that you wrote even execute if n or r are vectors? Even for scalar r and n, will it yield the correct answer? Try your own code. Think about what you see as a result.

And stop suggesting who should be marked the correct answer, yours or anyone.

Need I say it again? Answers is not a power struggle. It is not a place for those who would bully or control others into marking your answer as correct. It is not a race to see who can get every possible "point"awarded to you.

Answers is here to help the person who asks a question, or for the person who might look for an answer in the future. Please stop trying to make it into your narcissistic quest for glory.

[Jan]

John BG, try it:

r = rand(1,3);
N = 3:5;
% r and N are vectors. Now run your code.