{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Math 321 - The Dirichlet Test

# Math 321 - The Dirichlet Test - The Dirichlet Test...

This preview shows pages 1–2. Sign up to view the full content.

The Dirichlet Test Theorem (The Dirichlet Test) Let X be a metric space. If the functions f n : X C , g n : X IR , n IN obey F n ( x ) = n m =1 f m ( x ) is bounded uniformly in n and x g n +1 ( x ) g n ( x ) for all x X and n b g n ( x ) B n converges uniformly to zero on X then n =1 f n ( x ) g n ( x ) converges uniformly on X . Proof: The trick to this proof is the summation by parts formula, which we now derive. s n ( x ) = n s k =1 f k ( x ) g k ( x ) = F 1 ( x ) g 1 ( x ) + n s k =2 [ F k ( x ) F k - 1 ( x )] g k ( x ) = F 1 ( x ) g 1 ( x ) + n s k =2 F k ( x ) g k ( x ) n s k =2 F k - 1 ( x ) g k ( x ) = n s k =1 F k ( x ) g k ( x ) n - 1 s k =1 F k ( x ) g k +1 ( x ) = n s k =1 F k ( x ) ± g k ( x ) g k +1 ( x ) ² + g n +1 ( x ) F n ( x ) So if m > n , the diFerence between the m th and n th partial sums is s m ( x ) s n ( x ) = m s k = n +1 F k ( x ) ± g k ( x ) g k +1 ( x ) ² + g m +1 ( x ) F m ( x ) g n +1 ( x ) F n ( x ) If M = sup b | F n ( x ) | v v x X, n B , v v s m ( x ) s n ( x ) v v M m s k = n +1 ± g k ( x ) g k +1 ( x ) ² + Mg m +1 ( x ) + n +1 ( x ) = M ± g n +1 ( x ) g m +1 ( x ) ² + m +1 ( x ) + n +1 ( x ) = 2 n +1 ( x ) (1) since g m +1 ( x ) 0, g n +1 ( x )

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 3

Math 321 - The Dirichlet Test - The Dirichlet Test...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online