User talk:ZacharyS

Welcome! Glad you voted on the IDRIVE article. Please let me know if I can help you. Maurreen (talk) 07:48, 31 July 2005 (UTC)[reply]

IDRIVE

You showed support for This week's Improvement Drive.
This week Lhasa was selected to be improved to featured article status.
Hope you can help…

VMS

Hi, Zachary,

I read your very important comment about OpenVMS and decided to improve the article, so people like you could understand it better. I worked with VMS for many years, so it earned a special place in my heart :)

Can you please tell me how can i improve it? What interests you - the people who work with VMS, the use of VMS in different industries, its capabilities in comparison to other operating systems? Please tell me, and i will improve the article accordingly.

Also, can you please tell me how well do you understand the following concepts, which are important to understanding of VMS:

operating system
database
server
storage
high availability
dual redundancy

Thank you very much for your comments!--Amir E. Aharoni 06:50, 19 September 2005 (UTC)[reply]

Square-root, nth root by hands

Hi, Zachary,

Here my (complicate modulus) method to obtain the n-th root of any number:

Given a real number A, we would like to know the "a" that satify: a^n = A

You've to know that:

$a^{n}=\sum _{x=1}^{a}[x^{n}-(x-a)^{n}]$

I call x^n-(x-a)^n = Mn = complicate modulus, since it "cut" any A in "a" parts (that has different size)

So to have a simple rule to teach your young students:

Keep from Tartaglia's triangle for (x-1)^n the n row (equal to your n root)

knowing that the "magic" complicate modulus is: x^n-(x-a)^n

from tartaglia remove 1st element, and change the sign of all the others.

Now you just need to subtract from you A (for a times) the value of x^n-(x-a)^n , from 1 to a.

Example:

A=9 who is the square root of 9 ?

line n=2 of tartaglia fo (x-1) is: 1 -2 1

remove 1, change signe you've: M2 = 2x-1

tabulate:

x   M2       A-Mnx
              9
1   2x-1= 1  9-1=8
2   2x-1= 3  8-5=3
3   2x-1= 5  3-3=0  ok 3 is square root of 9 !

The process is slowly for big numbers, but easy works for any n !

And if you have an A that is a non power ? The same

Example:

A=9 who is the square root of 9 ?

line n=2 of tartaglia fo (x-1) is: 1 -2 1

remove 1, change signe you've: M2 = 2x-1

tabulate:

x   M2            A-Mnx
                  28
1   3x^2-3x+1= 1   28-1=27
2   3x^2-3x+1= 7   27-7 = 10
3   3x^2-3x+1= 19  10-19= R 1  OK 28 has non integer cubic root !

And if you wanna have the result with 1 decimale ?

Simple: for more precision, keep precision factor 1/k and use my new "step sum" :

 a^n = Sum from 1/k to a (step 1/k)  of 3x^2/k - 3x/k^2 + 1/k^3

For example k=10, precision 1/10

tabulate:

 M3 = 3x^2/10-3x/100+1/100 (from 0.1 step 0.1)

30

0.1	0.001	29.999
0.2	0.007	29.992
0.3	0.019	29.973
0.4	0.037	29.936
0.5	0.061	29.875
0.6	0.091	29.784
0.7	0.127	29.657
0.8	0.169	29.488
0.9	0.217	29.271
1	0.271	29
1.1	0.331	28.669
1.2	0.397	28.272
1.3	0.469	27.803
1.4	0.547	27.256
1.5	0.631	26.625
1.6	0.721	25.904
1.7	0.817	25.087
1.8	0.919	24.168
1.9	1.027	23.141
2	1.141	22
2.1	1.261	20.739
2.2	1.387	19.352
2.3	1.519	17.833
2.4	1.657	16.176
2.5	1.801	14.375
2.6	1.951	12.424
2.7	2.107	10.317
2.8	2.269	8.048
2.9	2.437	5.611
3	2.611	3
3.1	2.791	0.209   < stop at 3.1 that is what you're looking for
3.2	2.977	-2.768

if you wanna 1/100

 M3 = 3x^2/100-3x/10000+1/10000 (from 0.01 step 0.01)

Till the limit when k->infinite, than you've the integral from 0 to "a", of 3x^3... that is a^3 as you may need if you want the infinite precision root of a non n power number.

I know that from long time, but

...No way to have this officially advertised (anywhere !)

ciao Stefano robotec(at)netusrf.it — Preceding unsigned comment added by 94.81.217.96 (talk) 10:24, 29 September 2013 (UTC)[reply]