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1)
Since ( ) exists, then
( )
( )
( )
( ( ) ) ( )
( ) ( )
So we get: in case ( ) exists, it equals to 0.
Inductively continuing this statement, we get that the same holds for
Since ( )( ) exists, we get that
( )( )
Inductively getting back to lower derivatives we get:
( )( )
for
So the statement is proved.
2)
Consider a function
( ) | ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
|
( ) is a linear combination of functions ( ) ( ) ( ).
Since are continuous on [ ], ( ) is continuous on [ ].
Since are differentiable on ( ), ( ) is differentiable on ( ).
D O M Y A S S I G N M E N T SUBMIT
W W W . A S S I G N M E N T E X P E R T. C O M
Sample: Real Analysis - Continuity and Differentiability
1
( ) | ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
|
( ) | ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
|
Last 2 equalities hold because matrices the determinant is taken has 2 identical rows.
By Rolle’s theorem,
( ) ( )
( ) | ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
|
The statement is proved.
D O M Y A S S I G N M E N T SUBMIT
W W W . A S S I G N M E N T E X P E R T. C O M
2
3)
Consider a function
( )
( (
) ( ))
It is defined at *
+, continuous at *
+ (because f is continuous) and twice-
differentiable on (
). By mean value theorem,
(
) ( ) (
) ( )
( ) ( )
( ( ) (
) ( ) ( ))
( ) (
) ( )( )
( )
( (
) ( ))
Consider a function on *
+. It is continuous on this interval and
differentiable on (
) (because is twice differentiable). Thus, by mean value theorem,
(
) ( ) (
) ( )
Substituting this into expression we got earlier we get:
D O M Y A S S I G N M E N T SUBMIT
W W W . A S S I G N M E N T E X P E R T. C O M
3
( ) ( ) (
) ( )( )
( ) ( )
( ) (
) ( )
Since (
) ( ) we proved the needed statement.
1)
is continuous at [ ].
∫ ( )
Suppose
( )
for some [ ].
If we take | ( )|
in the definition of continuity of at point we get:
( ) ( ) | ( ) ( )| | ( )|
| ( )|
( ) ( )
| ( )|
So
| ( )| | ( )|
at in the interval ( ) ( ).
D O M Y A S S I G N M E N T SUBMIT
W W W . A S S I G N M E N T E X P E R T. C O M
4
Then
∫ ( )
∫ ( )
( ) ( ) ∫
| ( )|
( ) ( )
| ( )|
( ( ) ( ))
So we get contradiction with
∫ ( )
So ( ) at [ ].
D O M Y A S S I G N M E N T SUBMIT
W W W . A S S I G N M E N T E X P E R T. C O M
5