# ps6solns - Math 337 Winter 09 Problem set 6 solutions 8.3 B...

This preview shows pages 1–3. Sign up to view the full content.

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.

Unformatted text preview: Math 337 Winter 09, Problem set 6 solutions 8.3 B For x ∈ [0 ,n ] we have | f n ( x ) − e − x | = 0 . For x ∈ [ n,n + e n ] | f n ( x ) − e − x | ≤ | f n ( x ) | + e − x ≤ 2 e − n . For x ≥ n + e − n | f n ( x ) − e − x | = e − x ≤ e − n . Thus in all cases | f n ( x ) − e − x | ≤ 2 e − n . This shows that f n ( x ) goes to e − x as n → ∞ , uniformly on [0 , ∞ ). integraltext ∞ e − x dx = 1. On the other hand integraldisplay ∞ f n ( x ) = integraldisplay n e − x dx + integraldisplay n + e n n f n ( x ) = integraldisplay n e − x dx + 1 2 → 3 2 This does not contradict Theorem 8.3.1 since that theorem applies only to bounded intervals. 8.3 I Given ǫ > 0 choose n so that bardbl f n − f bardbl ≤ ǫ . (Here the norm refers to the sup norm on C [ a,b ]). This implies that | U ( f n ,P ) − U ( f,P ) | ≤ ǫ ( b − a ) for any partition P (supply a proof), and the same goes for lower sums. Since f n is Riemann integrable by Riemann’s integra- bility criterion we may find P so that | U ( f n ,P ) − L ( f n ,P ) | < ǫ . It follows by the triangle inequality that | U ( f,P ) − L ( f,P ) | ≤ | U ( f,P ) − U ( f n ,P ) | + | U ( f n ,P ) − L ( f n ,P ) | + | L ( f n ,P ) − L ( f,P ) | ≤ ǫ ( b − a ) + ǫ + ǫ ( b − a ) . Since this quantity can be made as small as we please by taking ǫ sufficiently small (alternately use ǫ (1 + 2( b − a )) − 1 in place of ǫ in the above argument) Riemann’s criterion is satisfied for f . 8.4 C Use calculus to verify that xe − x ≤ e − 1 for x ≥ 0. Thus x n e − nx ≤ e − n for x ∈ [0 , ∞ ). Since the geometric series ∑ e − n is summable the Weierstrass M-test implies that the series in question converges uni- formly on [0 , ∞ ). There is no need to show the uniform convergence on compact intervals separately. 1 2 8.4 J To have a visual image of what is going on here think of { f n ( k ) } as a two dimensional array of numbers indexed by k and n where n moves to the right as k increases and down as n increases. Thus the limits occur along each row and the sums are summations down each column. Needless to say the proof we are about to give in no way depends on the visual image, but is perhaps easier to understand if one has this image....
View Full Document

{[ snackBarMessage ]}

### Page1 / 5

ps6solns - Math 337 Winter 09 Problem set 6 solutions 8.3 B...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online