Advanced MathThis formula describes the multiplication rule for finite summation.
4 weeks 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.
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Best Answer
BerthaAdamsIsTaken Staff answered 4 weeks 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)
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