3 years ago
This formula describes the multiplication rule for finite summation.
(sum_(k=0)^(n)a_(k))sum_(j=0)^(n)b_(j)=sum_(k=0)^(n)sum_(j=0)^(n)a_(k)b_(j)
This formula is called Langrange's identity.
(sum_(k=1)^(n)a_(k)b_(k))^(2)=(sum_(k=1)^(n)a_(k)^(2))sum_(k=1)^(n)b_(k)^(2)-sum_(k=1)^(n)sum_(j=K+1)^(n)(a_(k)b_(j)-a_(j)b_(k))^(2)
Explain the difference between these two summation rules. Why is the product of two sums a double row and the square of the sum is a double row? Why does the square intersect the terms but not the product? Provide proof.
1 Answers
Best Answer
BerthaAdamsIsTaken Staff answered 3 years ago
BerthaAdamsIsTaken Staff answered 3 years ago
Step 1
1) Proof:
first thing is like (a_(0)+a_(1)+a_(2)+*s+a_(n))., (b_(0)+b_(1)+b_(2)+*s+b_(n))
=(a_(0)b_(0)+a_(1)b_(0)+*s+a_(0)b_(n))+
+(a-[1)b_(0)+*s+a_(1)b_(n))+*s+(a_(n)b_(0)+*s+a_(n)b_(n))
Step 2
2) Proof:
for this we have to split each term
lets take individual terms first LHS
(1) (sum_(k=1)^(n)a_(k)b_(k))^(2)=sum_(k=1)^(n)a_(k)^(2)b_(k)^(2)+2sum_(i=j)^(n-1)sum_(j=i+1)^(n)a_(i)b_(i)a_(j)b_(j)
(sum_(k=1)^(n)a_(k)b_(k))^(2)=sum_(k=1)^(n)a_(k)^(2)b_(k)^(2)+2sum_(i=1)^(n-1)sum_(j=i+1)^(n)a_(i)b_(i)a_(j)b_(j)
now lets take RHS
(2) (sum_(k=1)^(n)a_(k)^(2))(sum_(k=1)^(n)b_(k)^(2))=
sum_(i=1)^(n)sum_(j=1)^(n)a_(i)^(2)b_(j)^(2)
=sum_(k=1)^(n)a_(k)^(2)b_(k)^(2)+sum_(i=1)^(n-1)sum_(j=i+1)^(n)a_(i)^(2)b_(j)^(2)+sum_(j=1)^(n-1)sum_(i=j+1)^(n)a_(i)^(2)b_(j)^(2)
and
(3) sum_(i=1)^(n-1)sum_(j=i+1)^(n)(a_(i)b_(j)-a_(j)b_(i))^(2)
=sum_(i=1)^(n-1)sum_(j=i+1)^(n)a_(i)^(2)b_(j)^(2)+sum_(j=1)^(n-1)sum_(i=j+1)^(n)a_(i)^(2)b_(j)^(2)-2sum_(i=1)^(n-1)sum_(j=i+1)^(n)a_(i)b_(i)a_(j)b_(j)
such that, so from this we can write
(sum_(k=1)^(n)a_(k)b_(k))^(2)=(sum_(k=1)^(n)a_(k)^(2))sum_(k=1)^(n)b_(k)^(2)-sum_(k=1)^(n)sum_(j=k+1)^(n)(a_(k)b_(j)-a_(j)b_(k))^(2)