(1) problem
differentiation [intergration of sint/t], limit from 0 to x^2
let say f(t)=t, F(t)=t^2/2, if f(t)=sint/t, F(t)=intergration of sint/t
d/dx [intergration of sint/t] limit from 0 to x^2
d/dx [F(t)] limit from 0 to x^2
{d/dx [F(t)} = differentiation of F(t)
then {d/dx [F(t)} = [f(t)] limit from 0 to x^2
f(x^2)-f(0)
(sin x^2/x)-(sin0/0)
(sin x^2/x)-1
in this correct?
(2) problem
d/dt P(t)=P`(t)
t=0, P`(0)=0
t->infinite, then P`(infinite)=f(h)/h
in this right?
thanks a lot.
(2) problem is a part of question.
d/dt P(t)=P`(t)
t=0, P`(0)=0
t->infinite, then P`(infinite)=f(h)/h
in this right?
---End Quote---
I'm not quite sure what you're trying to do here...
(2) problem is only sub part of the full question that is to find the P(t) or Q(t) when t=0 and t->infinite
d/dt P(t) = -aP(t)+bQ(t)--1
d/dt Q(t) = aP(t)-bQ(t)--2
P(t)+Q(t) = 1
a and b are real number.
let d/dt P(t) = P`(t), d/dt Q(t) = Q`(t)
from the eq 1 and 2
Q(t)=[a-Q`(t)]/[a+b]---3
P(t)=[b-P`(t)]/[a+b]---4
from the eq 3 and 4
when t=0, P`(t)=0 or Q`(t)=0
Q(t)=[a-0]/[a+b]
P(t)=[b-0]/[a+b]
P(t)+Q(t) = 1, (a/[a+b])+(b/[a+b])=1 {prove}
thus when t=0, Q(t)=a/(a+b) or P(t)=b/(a+b)
but, i was wondering if t->infinite, what would be the value of P`(t)?
Could this [f(h+a)-f(a)]/h to be used?
lim t->infinite [P(h+infinte)-P(infinite)]/h
P(h)/h in this correct?
recall eq 4
P(t)=[b-P(h)/h ]/[a+b] will this be answer?