4-stage fixed-point binomial pipeline >>>>> urgent help

[dolly]

now i m going to explain it
(a + b)^n = a^n + n a^(n-1) b + n (n-1) a^(n-2) b^2 / 2 + ...

the value of n is fixed i.e n= 4 as it already mentioned in the objective
that it goes upto 4 stages the value or inputs for a and b is may also
fixed or given by me too say that i give the in stimulus of a =5 and for
b= 4
can any one please help me in making this ? i really need a urgent help
please mail me on shehnaz.tariq@gmail.com or uzmanamin@yahoo.com
i ll be very thankful

 

[Prasanth Kumar]

On Wed, 2004-10-27 at 15:08 -0400, dolly wrote:

now i m going to explain it
(a + b)^n = a^n + n a^(n-1) b + n (n-1) a^(n-2) b^2 / 2 + ...

the value of n is fixed i.e n= 4 as it already mentioned in the objective
that it goes upto 4 stages the value or inputs for a and b is may also
fixed or given by me too say that i give the in stimulus of a =5 and for
b= 4
can any one please help me in making this ? i really need a urgent help
please mail me on shehnaz.tariq@gmail.com or uzmanamin@yahoo.com
i ll be very thankful

Would it not be easier to add the numbers first and then square them
twice in the subsequent two stages?

 

[reddy]

The Problem is not clear........ Please be more specific.
If you explain the problem in detail, then somebody can try to
answer....... Sorry

-pradeep

 

[dolly]

hey
i jus want a code to implement binomial theorem to the first 4 steps in
this way n is fixed (n =4)
result = a^n + na^(n-1)b + n (n-1) a^(n-2) b^2 / 2 + n (n-1)(n-2)
a^(n-3) b^3 / 6.
i want this formula to be computed the values of a and b are the inputs
;obviously.
so please now any one can help me
thanking you

 

[dolly]

hey
i jus want a code to implement binomial theorem to the first 4 steps in
this way n is fixed (n =4)
result = a^n + na^(n-1)b + n (n-1) a^(n-2) b^2 / 2 + n (n-1)(n-2)
a^(n-3) b^3 / 6.
i want this formula to be computed the values of a and b are the inputs
;obviously.
so please now any one can help me
thanking you

 

[John_H]

"dolly" <shehnaz.tariq@gmail.com> wrote in message
news:61c4584eb08dccb59e34ec81ff47892f@localhost.talkaboutprogramming.com...

hey
i jus want a code to implement binomial theorem to the first 4 steps in
this way n is fixed (n =4)
result = a^n + na^(n-1)b + n (n-1) a^(n-2) b^2 / 2 + n (n-1)(n-2)
a^(n-3) b^3 / 6.
i want this formula to be computed the values of a and b are the inputs
;obviously.
so please now any one can help me
thanking you

Is the fixed "n" the n in your equation? Start with algebra.
result = a^4 + 4(a^3)b+6(a^2)(b^2)+6a(b^3)

What "we" ay not understand is why you want a binomial expansion if all you
want is (a+b)^4.
Add a+b. Multiply the result by itself. Multiply the new result by itself.
There ya go.

Engineering is results-oriented. Classes often do weird stuff to shed some
light on the subject. What your perspective is on trying to "implement"
this equation is unknown.

Obviously.

 

[dolly]

i got the clue but i m not geting how to code this equation ; this is the
binomial equation (a simple binomial theorem up to 4 stages and that whole
is done in pipeline style)
secondly i dont want a binomial expansion
i just want to solve the equation of binomail theorem upto 4 stages i
wrote that 4 stages!!!

 

[Raghavendra]

"dolly" <shehnaz.tariq@gmail.com> wrote in message news:<bee39c29396644c5c86d8616f4ede8df@localhost.talkaboutprogramming.com>...

hey
i jus want a code to implement binomial theorem to the first 4 steps in
this way n is fixed (n =4)
result = a^n + na^(n-1)b + n (n-1) a^(n-2) b^2 / 2 + n (n-1)(n-2)
a^(n-3) b^3 / 6.
i want this formula to be computed the values of a and b are the inputs
;obviously.
so please now any one can help me
thanking you

hi,
In HDL implemeting this equation is not straightforward as it is in
C programming.Here u need to have adders,multipliers..then wire them
to get the results.

Raghavendra.S