Looking for an identity or process to break one function into to

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Would like to go from here:
(A'*C - B*C) / (A + B) to
A'*C/A - B*C/(A*(A+B))
Can anyone name this identity or process which pulls A'*C/A out of (A'*C - B*C) / (A + B) as shown above?
Thank you.
  1 Comment
CD
CD on 8 Feb 2019
I should have been more clear. I forgot to point out that there is a value "A" and value "A' " where A' = 1 -A.
I've tested this in excel:
A A' B C
0.2 0.8 2 3
(A'C - BC) / (A + B)
-1.636
A'C/A - BC/(A(A+B)
-1.636

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Answers (1)

Walter Roberson
Walter Roberson on 8 Feb 2019
False. Suppose A = 5 and B = 11 then
>> syms C
>> (5*C - 11*C)/(5+11)
ans =
-(3*C)/8
>> 5*C/5 - 11*C/(5*(5+11))
ans =
(69*C)/80
Not even the same sign.
  1 Comment
Walter Roberson
Walter Roberson on 8 Feb 2019
Consider
(Ap*C - B*C)/(A + B) = Ap*C/A - B*C/(A*(A + B))
multiply both sides by A/C to get
(Ap*C - B*C)*A/((A + B)*C) = (Ap*C/A - B*C/(A*(A + B)))*A/C
On both sides, the C cancel in the top and bottom. On the right side th A cancel on the top and bottom
(Ap - B)*A/(A + B) = Ap - B/(A + B)
normalize the right side into a fraction
(Ap - B)*A/(A + B) = (A*Ap + Ap*B - B)/(A + B)
discard the denominator
(Ap - B) * A = A*Ap + Ap*B - B
expand
Ap*A - A*B = A*Ap + Ap*B - B
cancel Ap*A on both sides:
-A*B = Ap*B - B
factor right side:
-A*B = (Ap-1)*B
Ap is 1-A so (Ap-1)*B is (1-A-1)*B = -A*B which is the left hand side. Therefore the two sides are equal -- except for the cases where A=-B or A = 0 or C = 0, which would have to be examined more carefully to avoid multiplication or division by 0 giving false equations.

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