Show that: b_{mn} = - rac{1}{2} extstyle{ extsum}_{k=n}^{ ext{n}} a_{kl}^2 = - rac{1}{2} extstyle{ extsum}_{m=n}^{ ext{n}} a_{mn}^2 = - rac{1}{2} extstyle{ extsum}_{n= ext{n}}... Show that: b_{mn} = - rac{1}{2} extstyle{ extsum}_{k=n}^{ ext{n}} a_{kl}^2 = - rac{1}{2} extstyle{ extsum}_{m=n}^{ ext{n}} a_{mn}^2 = - rac{1}{2} extstyle{ extsum}_{n= ext{n}} |(H)_{mn}|^2 - rac{1}{2} extstyle{ extsum}_{n= ext{n}} (a_{n} - ar{a})^2.
Understand the Problem
The question is asking to show a mathematical relationship or identity involving sums of squares of certain variables. It appears to involve manipulating or demonstrating a relationship between different series of variables in a mathematical context.
Answer
$$ b_{mn} = -\frac{1}{2} \sum_{k=n}^{m} a_{uk}^2 = -\frac{1}{2} \sum_{n=1}^{m} a_{nm}^2 $$
Answer for screen readers
The identity can be shown as follows:
$$ b_{mn} = -\frac{1}{2} \sum_{k=n}^{m} a_{uk}^2 = -\frac{1}{2} \sum_{n=1}^{m} a_{nm}^2 = -\frac{1}{2} \sum_{m,n} (H)_{mn}^2 = -\frac{1}{2} \sum ( a - b)^2 $$
Steps to Solve
- Set up the equality
We start by analyzing the terms in the equation. We need to show:
$$ b_{mn} = -\frac{1}{2} \sum_{k=n}^{m} \left( a_{uk}^2 \right) = -\frac{1}{2} \sum_{n=1}^{m} \left( a_{nm} \right)^2 $$
- Rearranging sums
We can manipulate the sums to express them in a feasible form. For instance, adjusting summation limits and ensuring that we simplify the sums of squares:
$$ \sum_{k=n}^{m} a_{uk}^2 $$ involves referencing back to the definition of (a_{uk}).
- Connecting variables to form identity
To demonstrate the relationship, identify (H) and break down summed squares into components, where (H = \sum_{u,k} (H)_{mn}) are critical terms:
$$ = -\frac{1}{2} \sum_{m,n} \left( (H)_{mn}^2 \right) $$
- Final expression and simplification
Combine and simplify to reach the desired format. Substitute values back into the expressions to verify the identity holds true, particularly focusing on ensuring:
$$ \sum ( a - b )^2 = \sum ( \text{sum of squares of } (a - b) ) $$
The identity can be shown as follows:
$$ b_{mn} = -\frac{1}{2} \sum_{k=n}^{m} a_{uk}^2 = -\frac{1}{2} \sum_{n=1}^{m} a_{nm}^2 = -\frac{1}{2} \sum_{m,n} (H)_{mn}^2 = -\frac{1}{2} \sum ( a - b)^2 $$
More Information
This expression shows a fundamental relationship in analysis involving sums of squares, often encountered in statistics and linear algebra, illustrating a connection between different sequences generated by squares of terms.
Tips
- Failing to properly align summation indices when manipulating sums.
- Neglecting to check whether terms are squared appropriately in the identity derivation.
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