sum of digit

phrontister · 2012-05-09 01:29:57

Thanks. Done.

I'd better say goodnight. Very early start at work in the morning.

Last edited by phrontister (2012-05-09 01:33:00)

bobbym · 2012-05-09 01:33:34

Try

Sqrt[2.] Do not forget the period after the 2

phrontister · 2012-05-10 01:38:38

Yes, I got the extra decimal places. Without the period the display showed the square root symbol.

This gave an incorrect result: Sqrt[2*3]*Sqrt[2*3] - Sqrt[2.*3]*Sqrt[2.*3]
This gave a correct result: Sqrt[2.*3]*Sqrt[2*3] - Sqrt[2.*3]*Sqrt[2.*3]
This one's interesting: N[Sqrt[2.*3].Sqrt[2*3] - Sqrt[2.*3].Sqrt[2.*3], 20]

W|A yielded zero for all.

I was just playing around with the period, the dot and the '*' to see what different effects they had.

Last edited by phrontister (2012-05-10 01:47:49)

bobbym · 2012-05-10 01:48:52

The problem I posed is designed to illustrate a technique. It is most effective against calculators. M tries to compensate for roundoff errors and usually does a good job.

phrontister · 2012-05-10 02:02:29

My answer for Sqrt[2.] is 1.4142135623730951, which is the same as Windows Calculator.

It is most effective against calculators.

I'm not sure what you mean by that.

bobbym · 2012-05-10 02:07:08

Hi;

I am talking about the problem with the subtraction of two square roots. Most calculators unless they have extreme amounts of digits will give an answer of zero to that problem. Even M if used incorrectly will give 0.

phrontister · 2012-05-10 02:31:12

bobbym wrote:

Try
Sqrt[2.] Do not forget the period after the 2

Did you mean that I should try that in the difference between the two square roots problem?

My answer for N[Sqrt[2+1/10000000000000000]-Sqrt[2.],50] is 0.

But N[Sqrt[2+1/10000000000000000]-Sqrt[2],50] =
3.5355339059327375778100479863650260262257526608698*10^-17

Why is that?

Also, if I use Sqrt[3.] instead of Sqrt[3] I get a different answer (not zero) than the one you gave. But Sqrt[2.] results in 0. (???)

bobbym · 2012-05-10 02:40:26

Hi;

Did you mean that I should try that in the difference between the two square roots problem?

No, that was meant to test whether you had adjusted your display precision correctly.

The first problem has 16-17 digits of precision. That is not enough to overcome the instability of the question. When you force M to work to 50 digits of precision he was able to get the correct answer.

phrontister · 2012-05-10 02:59:54

Hi Bobby,

W|A gives the right answers even without specifying precision.

When you force M to work to 50 digits of precision he was able to get the correct answer.

But not if I include the period (in my M7). Using 100 digits of precision:

...with Sqrt[3.] instead of Sqrt[3] I get 2.220446049250313*10^-16 instead of 2.88675134594812879848948268627538082462782983934251295103889170779953\
8687162958829196014249435742169*10^-17.

...and with Sqrt[2.] instead of Sqrt[2] I get zero instead of 3.53553390593273757781004798636502602622575266086977757726046890388419\
6341402064375958344406012265096*10^-17.

bobbym · 2012-05-10 03:07:27

Alpha is doing it all automatically for you but it is much less reliable with its answers.

When you put in 2 you have an exact number. When you input 2. you have a floating point number which is just an approximation.

phrontister · 2012-05-10 03:20:16

Yes, I thought it was something like that. Not important with most of what I'd ever do.

Thanks, Bobby. Bed now...

bobbym · 2012-05-10 03:25:24

Not important with most of what I'd ever do.

Could be. But at least you have been exposed to it.

phrontister · 2012-05-10 03:32:56

Yes, thanks for doing that.

Given the degree of precision required in certain fields and that M is so widely used in them (I think), the fact that these anomalies occur is actually quite interesting.

One day everything will be perfect...if we don't blow ourselves into smithereens first.

bobbym · 2012-05-10 03:36:44

Hi phrontister;

These are not exactly anomalies. They are inherent in the way computers
do arithmetic. This is not well known at all but computers can not divide or subtract. They have trouble multiplying! This is not a software problem.

Have you forgotten a piece of work we did together on using subtraction in LB for a loop counter? That was an example of the difficulties computers have doing arithmetic.

phrontister · 2012-05-10 10:18:35

Good morning, Bobby...time for just this one.

Yes, I do recall that LB loop counter work. That was over two and a half years ago! I'd been a MIF member for only 2 months by then. How time flies when you're having fun! I re-read the old posts.

I hadn't realised until then that computers had difficulties with maths...even more difficulties than I do! And now you've shown they struggle in other areas.

Well, that just proves their human influence (see my signature).

Must go...catch you later.

bobbym · 2012-05-10 10:22:59

Hi phrontister;

I guess man made computers, man makes mistakes, computers make mistakes.

anonimnystefy · 2012-05-13 07:53:20

phrontister wrote:

Everything in the second set of brackets.
I don't know what this does:
Nor this:
I've seen that funny symbol in other posts but I've never tried to work out what it means.

Hi phro

I remembered you like procedural programming so here is your explanation:

s=0;
for i=1 to n do
s=s+d[i];
end;
return s;

And

is the same as:

phrontister · 2012-05-14 00:21:06

Thanks for that, stefy.

It gave me a pretty good understanding of gAr's formula (I think).

I entered the formula into Excel (see image) and I also wrote a little program for it in LibertyBASIC that seems to work fine.

I looked up "sigma summation" on the net to get some more ideas (and MIF has a page on it too), but I'm not ready to travel very far down that road yet.

anonimnystefy · 2012-05-14 00:22:43

You're welcome.

bobbym · 2012-05-14 00:26:21

Hi phrontister;

gAr's formula is made for a CAS rather than Excel.

phrontister · 2012-05-14 02:51:56

Hi Bobby,

Yes, I thought it probably would be, but my mind is on other things at the moment and as it was easy enough to do in Excel I thought I'd at least do it there.

Column A's digits can be deleted and replaced by a choice of digits. As each digit is entered (starting from A3), the adjacent column B cells display the sum of all permutations for the numbers entered...and the other cells also update.

'Conditional formatting' prevents cells from displaying error messages.

Last edited by phrontister (2012-05-14 02:57:53)

bobbym · 2012-05-14 04:30:25

Hi phrontister;

but I'm not ready to travel very far down that road yet.

Supposing there was a number 1357982. To use sigma or summation you
just type:

Sum[i,{i,{1,3,5,7,9,8,2}}]

answer will be 35. Great thing about a CAS, you learn as you go.

juantheron · 2012-05-14 15:43:10

Thanks member for heavy and Nice discussion

bobbym · 2012-05-14 21:58:30

Hi juantheron;

Most of it was about things not directly related to your question. But it is where your question seemed to go.

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