C# loop on fraction

Question

I need someone to give me an idea on how to go on about this problem.Using a loop to calculate the fraction , There is no common value.I want to get the sum

Eg for fraction :

1  1/5  1/10  1/15  1/20 … 1/290  1/295  1/300

code snippet:-

int sum=0;
for(int i=1;i<=60 ;i++)
{
   int sum=1
}

You want to get the sum of `1/(5*i)` for `i` in `1:60`? You know that's a geometric series, which means there's a [formula](https://en.wikipedia.org/wiki/Geometric_series#Formula) to get the result instantly — Rafalon, May 31 '18 at 09:14
To start with, the sum is not an integer, so you probably want a `float`. Beyond that, it's not clear what the question is. — John Perry, May 31 '18 at 09:16
@JohnPerry I would suggest decimal for such accurate values instead of float — Ipsit Gaur, May 31 '18 at 09:18
@IpsitGaur Fair enough; I'm not that familiar with C#, but in any case there's a problem with `int sum`. — John Perry, May 31 '18 at 09:19
@IpsitGaur: What makes you think that a decimal type can store the terms more accurately than a `double`? — Bathsheba, May 31 '18 at 09:19
@Rafalon - it's not geometric! Geometric series' denominators increase exponentially. This is increasing linearly. — Alex Reinking, May 31 '18 at 09:37
@AlexReinking that's what I realized 3 minutes ago as my comment on Bathsheba's answer demonstrates. But hey, thanks for pointing it out anyway — Rafalon, May 31 '18 at 09:38

Bathsheba · Accepted Answer · 2018-05-31T09:53:31.143

5

These sort of problems are actually surprisingly non-trivial due to issues with working with floating point, and decimal types for that matter.

Accepting that you want a loop solution for this (a closed form solution for n terms does exist), first note that your series can be written as

1 + 1/5(1 + 1/2 + 1/3 + ... + 1/60)

Then note that a good rule of thumb when working with floating point types is to add the small terms first.

So an algorithm would be of the form

double sum = 0.0;
for (int i = 60; i >= 1; --i){
    sum += 1.0 / i;
}
sum = sum / 5 + 1;

Note the 1.0 in the numerator; that's there to defeat integer division.

Reference: Is floating point math broken?

edited May 31 '18 at 09:53

answered May 31 '18 at 09:22

Bathsheba

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Or `double sum = (1-Math.Pow(0.2,60))/(1-0.2);` right? – Rafalon May 31 '18 at 09:23
@Rafalon: Or just write out the answer as a constant! I think the OP wants a loop. – Bathsheba May 31 '18 at 09:24
You mean `sum / 5 + 1` – Alex Reinking May 31 '18 at 09:25
@Rafalon, yes that **could** be the thing what if the question state to do it with a loop ? – Drag and Drop May 31 '18 at 09:25
@DragandDrop & Bathsheba, that's right, I missed the "*Using a loop*" part – Rafalon May 31 '18 at 09:26
@AlexReinking: No I don't, unless my rewritten form is incorrect. – Bathsheba May 31 '18 at 09:26
@Rafalon, this should have been an answer btw ^^. beening able to out side of the box seeing the general pattern is a good thing and will make a great answer with little code and little explanation. And We like to see math use here. To render a non trivial issue trivial. So we can even compare the result of both. – Drag and Drop May 31 '18 at 09:26
@Bathsheba - It _is_ incorrect. OP wanted `1 + 1/5 + 1/10 + ... + 1/300`. You gave `5 + 1/5 * (1 + 1/2 + ... + 1/60) = 5 + 1/5 + 1/10 + ... + 1/300`. – Alex Reinking May 31 '18 at 09:27
1

@AlexReinking: Oops. Fixed. Let's publish jointly. And I'll upvote your answer for good feelings! – Bathsheba May 31 '18 at 09:28
@Bathsheba - works for me – Alex Reinking May 31 '18 at 09:31
@DragandDrop I was on my way of writing that, but I realized I made a mistake, it's not geometric, as you do not multiply by `0.2` at every step as I thought, this would have been `1 + 1/5 + 1/25 + 1/75 ...` – Rafalon May 31 '18 at 09:34

Alex Reinking · Answer 2 · 2018-05-31T09:41:32.300

3

͏Since you asked for a hint:

float sum = 1.0;
for (int i = 5; i <= ??; i += ??) {
    sum += 1.0/i;
}

What goes in place of the ??s?

edited May 31 '18 at 09:41

answered May 31 '18 at 09:19

Alex Reinking

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It's naughty to add the large terms first when working with floating point. – Bathsheba May 31 '18 at 09:20
@Bathsheba - It doesn't matter one iota in this case. The relative error between the two directions is `2.27e-16` – Alex Reinking May 31 '18 at 09:26
But it's still bad practice. If this finds its way into a language where the `double` allows you freedom to mess about with the exponent and mantissa, it *could* matter. – Bathsheba May 31 '18 at 09:27
1

I'm not really looking to discuss the finer points of floating-point best practice when OP is having trouble forming the loop in the first place. But you raise a fair point. – Alex Reinking May 31 '18 at 09:33
@Bathsheba interesting... Could you explain "why" this is a bad practice? Or give some reference to read, I would like to learn more about that :) – Damien Flury May 31 '18 at 09:43
1

^ Floating point is most accurate when adding numbers of similar magnitude. In floating point `1 + 1e-16` is exactly `1`, but you can actually express `1e-16 + 1e-16`. If you have a lot of small values, they could add up to a bigger one, which would be more accurate if the terms of your sum have a high dynamic range. – Alex Reinking May 31 '18 at 09:47
1

@DamienFlury: in floating point, adding a very small number to a very big number can be a no-op. So you want to give the small numbers chance in a sum of contributing to the final total. Therefore add them first. – Bathsheba May 31 '18 at 09:48
@AlexReinking and Bathsheba Alright, I think this makes sense, thank you! – Damien Flury May 31 '18 at 09:58

score 1 · Answer 3 · answered May 31 '18 at 09:23

1

try this code:

    double sum=1;
    for(int i=5; i<=300; i+=5)
        sum += (double) 1 / i;

The value of sum will be 1.93597408259035

answered May 31 '18 at 09:23

Saeed Bolhasani

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C# loop on fraction

3 Answers3