Which we called ” Omari’s formulas
“. And the coefficients:
are ” Omari’s coefficients “.
Some examples for calculating Omari’s coefficients:
Example 1:
=576
Example 2:
The next tables give the values of Omari’s coefficients for the integer n up to 5.
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Ank |
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k |
0 |
1 |
2 |
3 |
4 |
5 |
|
n |
2n+1 |
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0 |
1 |
1 |
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1 |
3 |
3 |
4 |
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2 |
5 |
5 |
20 |
16 |
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|
| 3 |
7 |
7 |
56 |
112 |
64 |
|
|
|
4 |
9 |
9 |
120 |
432 |
576 |
256 |
|
| 5 | 11 |
11 |
220 |
1232 |
2816 |
2816 | 1024 |
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Bnk |
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k |
0 |
1 |
2 |
3 |
4 |
5 |
|
n |
2n+1 |
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| 0 |
1 |
1 |
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| 1 |
3 |
1 |
4 |
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2 |
5 |
1 |
12 |
16 |
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|
| 3 | 7 |
1 |
24 |
80 |
64 |
|
|
| 4 | 9 | 1 |
40 |
240 |
448 |
256 |
|
|
5 |
11 | 1 |
60 |
560 |
1792 | 2304 |
1024 |
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Cnk |
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k |
0 |
1 |
2 |
3 |
4 |
5 |
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n |
2n |
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|
1 |
2 |
2 |
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2 |
4 | 4 | 8 |
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3 |
6 |
6 |
32 |
32 |
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4 |
8 |
8 |
80 |
192 |
128 |
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5 |
10 |
10 |
160 |
672 |
1024 |
512 |
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Dnk |
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k |
0 | 1 |
2 |
3 |
4 |
5 |
|
n |
2n |
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|
1 |
2 |
1 |
2 |
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|
2 |
4 |
1 |
8 |
8 |
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|
|
3 |
6 | 1 |
18 |
48 |
32 |
|
|
|
4 |
8 |
1 |
32 | 160 |
256 |
128 |
|
|
5 |
10 |
1 |
50 |
400 |
1120 |
1280 |
512 |
How to get Omari’s formulas
We will start from Moivere’s formula:
………..(9)
To define the last term in (9) if it is real or imaginary, we must define n is
it even or odd. Therefore, we will use 2n for even, and 2n+1
it even or odd. Therefore, we will use 2n for even, and 2n+1
for odd. We will
get two formulas:
get two formulas:
…(10 )
…………(11)
After separating real parts from imaginary, formulas (10) and (11) will give
four formulas. We will continue with one of them. The real part of formula (10)
is:
four formulas. We will continue with one of them. The real part of formula (10)
is:
……..…..(12 )
This formula is already known to experts, as will as the other three. Our goal
is to have in the right side terms with sinx only or cosx only. Therefore, we
need to eliminate sinx once, and cosx in the second time using the identity:
,
so we have eight formulas totally. Here, we will eliminate sinx:
is to have in the right side terms with sinx only or cosx only. Therefore, we
need to eliminate sinx once, and cosx in the second time using the identity:
,so we have eight formulas totally. Here, we will eliminate sinx:
.…………….(13)
Now, we have to rearrange terms so as to put together terms with the same
exponent of cosx:
exponent of cosx:
………..…..(8)
we got formula (8). The other formulas (1),…(7) could be got in the same way.
These formulas, in the first time, were put in empirical way, and were provedwith means of “mathematical induction” method; later, they were developed from
known mathematical formulas, as we saw. So, the proof with means of “mathematical
induction” exists.
One of uses of these formulas, we can see if we try to solve integrals, for example, sin(nx) or cos(nx) with sinx or cosx, such as cos(2n)x*sinx.
Eng. Zuhair Omari
Regestered in Jordan, at The Ministry Of Culture - Department Of National Library, under No.863/2008, March 25. 2008.
