How do I calculate  ( 2000)+1720+1641+1563+1486 +1410  = 9820?

            ( 2000) base number
            +1720
            +1641
            +1563
            +1486
            +1410
    -------------
Total  =  9820

1720 =  2000 - 80
1641 =  1720 - 79
1563 =  1641 - 78
1486 =  1563 - 77
1410 =  1486 - 76

How does this polynomial work?

The name of this polynomial?

One website reference with one similar example?

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Clarification of Question by guyyy-ga on 25 Feb 2006 14:37 PST

I forgot the word Please 
Please website references with similar examples will be appreciated?
Guy

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Request for Question Clarification by pafalafa-ga on 25 Feb 2006 14:51 PST

I'm not really sure what you're asking.  This looks like plain old addition to me:


://www.google.com/search?sourceid=navclient&ie=UTF-8&rls=GGLD,GGLD:2003-43,GGLD:en&q=%28+2000%29%2B1720%2B1641%2B1563%2B1486+%2B1410%3D


Can you elaborate?

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Clarification of Question by guyyy-ga on 25 Feb 2006 15:39 PST

Hello   pafalafa-ga 
>>  I'm not really sure what you're asking.
I do believe that.
Without sounding rude, I believe you have no idea what a polynomial is
otherwise you'll not make such a remark after all the explanations I gave. 

>>  This looks like plain old addition to me:
You did not read my question. [full stop]
You just guessed my question

>> Can you elaborate?
Did, you  read the word "polynomial" in my question?

Did you read my question? " How does this polynomial work?"

If you are working with math you'll know exactly I am after the
formula for this particular polynomial?

The formula which satisfies the result 9820 starting at 2000 and after
5 additions and
the same formula starting again at 2000 after 18 moves with the result 21689?
What's the mathematical formula?

This formula in expended form is "probably " like this
IMPORTANT --> (n+1)t_a - n(n+1)(2n+1)..... etc 
(I am saying it is "probably"like  this)
 
This problem can be built using a Pascal triangle.
A Pascal triangle is not the mathematical formula I am looking for.
It probably will help to understand the construction of the formula I
am looking for.

I am NOT after an algorithm ( I already got it). 
I am after the mathematical formula which probably has 3+ variables?

Guy

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Clarification of Question by guyyy-ga on 25 Feb 2006 19:02 PST

>>brix24-ga on 25 Feb 2006 17:57 PST..
>>Did you mean your sequence to be:

>>2000
>>1920 (rather than 1720) brix24-ga -there is no mistake 1720 is correct.
>>1841
>>etc?

brix24-ga, All I can say is "thank you", you understand most of my question.

The base is 2000 then the folowing numbers are
            1720
            1641
            1563
            1486
            1410
This is how it works, I constantly subtract n-1 to b and add it to the result.

Here is how I do it in a basic program
 T1=1800 :CN =81 : T2=2000
       PRINT "x1=";X1;"   CN=";CN; "   T1="; T1 ;"   T2=";T2
          FOR X1  = 1 TO 20
            CN=CN-1
               T1 = T1-CN
                T2= T1+T2
            PRINT "x1=";X1;"   CN=";CN; "   T1="; T1 ;"   T2=";T2
          NEXT X1       

How do I transform this program above into a mathematical formula?

In my "real" program T2 = 200 000 000+  (above 200 000000) requires
enormous quantities
of iterations to calculate a specific value.
This is the reason why I have presented some small numbers in my
example and want to transform this
algorithm into a simple math formula.

Big numbers tend to turn people off

brix24-ga you are very close.
I can follow your argument very well.
Just like you I get stack. We'll get it a way or the other.

Can you help me or anyone else to formulate my problem in a more
mathematical way so more mathematicians will understand?

I agree my problem is posed from a computer programmer point of view
and a mathematician point of view would be better.

Guy

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Clarification of Question by guyyy-ga on 25 Feb 2006 19:31 PST

>>myoarin-ga on 25 Feb 2006 19:10 PST     
>>Guyyy-ga,
>> As Brix pointed out, you have some kind of an error in your numbers.  
>>1720 does NOT equal 2000 - 80.

Thanks myoarin-ga for pointing this out
 
You are right,I made typo, the base should be 1800 and not 2000 

Sorry about the typo.

Guy

guyyy-ga, it appears that it is YOU who does not know what a
polynomial is.  See here: http://en.wikipedia.org/wiki/Polynomial.

Are you just making things up?  Or complicating things that should be simple?

Did you mean your sequence to be:

2000
1920 (rather than 1720)
1841
etc?

I take it you want a formula for the sum of n terms of this series.

You may have what I already have, e.g., for 6 terms, the formula is

n * (base number) - (starting difference, =80 here)* n * (n-1) / 2 +
(n-2)*1 + (n-3)*2 .... + 2*(n-3) + 1*(n-2)

Unfortunately, my math is rusty enough that I don't know the
simplified formula (n-2)*1 + (n-3)*2 .... + 2*(n-3) + 1*(n-2)

For 6 terms, the formula I have gives:

6*2000 -80*6*5/2 +4*1 + 3*2 +2*3 +1*4
=12,000 - 1,200 + 4 + 6 + 6 + 4
=10,820

I think what you want is something with three terms:

  n * (base number) 
- (starting difference)* n * (n-1) / 2 
+ (a simple formula in place of what I have: the (n-2)*1 + (n-3)*2 .....)

or
 n*BN - d*N*(n-1)/2 + (some simple term/formula)

For 7 terms, I get 

7*2000 - 80 * 7 * 6 / 2 + (5 + 4*2 + 3*3 + 2*4 + 5)
= 14,000 - 1680 + 35
= 12,355

Sorry, but I don't know the formula for simplifying the last part.

nelson-ga said:
>>>Are you just making things up?

No nelson-ga I am not making things up.

nelson-ga if you know what a polynomial is, please explain this 
polynomial:  S p(x) = 961n/6+32n^2-n^3/6 

nelson-ga if you can explain what this polynomial does I may listen to
your argument.

>>>Or complicating things that should be simple?

nelson-ga , what is the purpose of such remark?

nelson-ga this is about transforming a non linear address into a
linear address using polynomial?
Would like to hear more ....?

Don't tell me, just show me.

I am waiting

For you nelson-ga I am posing the problem in a different way.

This is the same problem.
What is the formula which satisfies?
( 2000)+n1+n2+n3+n4+n5  = 9820
and
( 2000)+n1+n2+n3....n17+n18 = 21689

Is that better that way?

nelson-ga, can you seriously help or are you just playing around?

Guy

Guyyy-ga,
On a two dollar question that looks like homework (which should not be
answered, see FAQs) you may have to put up with some off-hand
comments.

As Brix pointed out, you have some kind of an error in your numbers.  
1720 does NOT equal 2000 - 80.

"This is the same problem.
What is the formula which satisfies?
( 2000)+n1+n2+n3+n4+n5  = 9820"

There can be no formula to satisfy this since a negative value of ANY
size can be given to any "n", which is balanced in the addition by
positive values of the other "n"s to make the total 9820.

Get off your high horse.  You asked the question.  If you didn't
notice your typing (?) error when you reviewed your text before
posting it or when posting your clarification and comment, even after
Brix called attention to it, you cannot expect much respect.

For a $2.00 question that is written so opaquely, I don't think you
should be insulting the people trying to answer it. I too was
hopelessly confused, and I'm quite certain pafalafa knows what a
polynomial is.

Here is the fact you need to know to solve your problem. The sum of
the numbers 1 to N is just:

S(N) = 1+2+3+...+N = N(N+1)/2

Thus, the sum of the numbers N to K is

S(K) - S(N-1) = K(K+1)/2 - N(N-1)/2 = (K^2 + K - N^2 + N)/2


This tells you how to sum an