1.Let k >= 1 be an integer, and define Cn = SIGMA (1/(n+I)), from i=1 to kn
(a)Prove that {Cn} converges by showing it is monotonic and bounded.
(b)Evaluate LIMIT (Cn) AS n approach to the infinity
2. Let f(x) be integrable on [a,b], and let g(x) be nondecreasing and
continuously differentiable on [a,b]. Let (p be the element of P) be
a partition of [a,b], and define
U(f,g,p) = SIGMA (Mi(g(X's ith term )-g(X's i-1 term))) from i=1 to n
L(f,g,p) = SIGMA (mi(g(X's ith term )-g(X's i-1 term))) from i=1 to n
Use mean value theorem to prove that (inf U(f,g,p) such that p is the
element of P ) = (sup L(f,g,p) such that p is the element of P ) =
(INTEGRAL f(x)g?(x)dx, from a to b)
3. Given that ( INTEGRAL ln square(x)dx, from n to n+1 ) = ( INTEGRAL
ln square (n+x)dx, from 0 to 1 ) = ( INTEGRAL [[ln(n+x) ? ln(x) +
ln(n)]square] dx, from 0 to 1 ),
(a) Verify that (LIMIT [(n/ln(n))*[INTEGRAL ln square (x)dx ? ln
square (n)]] AS n -> the infinity ) = 1
(b) Compute (LIMIT ((n square)/ln(n))*[ INTEGRAL ln square (x)dx - ln
square (n) - ((ln(n))/n)] AS n -> the infinity ) |
Clarification of Question by
aoyama-ga
on
08 Dec 2003 16:41 PST
1.Let k >= 1 be an integer, and define Cn = SIGMA (1/(n+I)), from i=1 to kn
(a)Prove that {Cn} converges by showing it is monotonic and bounded.
(b)Evaluate LIMIT (Cn) AS n approach to the infinity
2. Let f(x) be integrable on [a,b], and let g(x) be nondecreasing and
continuously differentiable on [a,b]. Let (p be the element of P) be
a partition of [a,b], and define
U(f,g,p) = SIGMA (Mi(g(X's ith term )-g(X's i-1 term))) from i=1 to n
L(f,g,p) = SIGMA (mi(g(X's ith term )-g(X's i-1 term))) from i=1 to n
Use mean value theorem to prove that (inf U(f,g,p) such that p is the
element of P ) = (sup L(f,g,p) such that p is the element of P ) =
(INTEGRAL f(x)g?(x)dx, from a to b)
3. Given that ( INTEGRAL ln square(x)dx, from n to n+1 ) = ( INTEGRAL
ln square (n+x)dx, from 0 to 1 ) = ( INTEGRAL [[ln(n+x)-ln(x)+
ln(n)]square] dx, from 0 to 1 ),
(a) Verify that (LIMIT [(n/ln(n))*[INTEGRAL ln square (x)dx-ln
square (n)]] AS n -> the infinity ) = 1
(b) Compute (LIMIT ((n square)/ln(n))*[ INTEGRAL ln square (x)dx - ln
square (n) - ((ln(n))/n)] AS n -> the infinity )
|
