Posted by Sarita on .
A) How do you prove that if 0(<or=)x(<or=)10, then 0(<or=)sqrt(x+1)(<or=)10?
B) So once that is found, then how can you prove that if 0(<or=)u(<or=)v(<or=)10, then 0(<or=)sqrt(u+1)(<or=)sqrt(v+1)(<or=)10?
* calculus  Damon, Saturday, April 3, 2010 at 3:35pm
How do you prove that if 0(<or=)x(<or=)10, then 0(<or=)sqrt(x+1)(<or=)10?
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does the square root increase (is the derivative positive) as x goes from 0 to 10 ?
If so the left side of the domain is minimum and the right side is maximum of the function and we only need to test the ends.
d (x+1)^.5 / dx = .5 /sqrt(x+1)
that is positive everywhere in the domain so all we have to prove is the end points.
0 </= x </= 10
if x = 0
sqrt x+1 = sqrt 1 = 1
if x = 10
sqrt x+1 = sqrt 11 = 3.32
so
1 </ sqrt(x+1) </= 3.32
* calculus  Damon, Saturday, April 3, 2010 at 3:37pm
for part b again the derivative is positive throughout the domain so if v is right of u then sqrt (1+v) > sqrt(1+u)
* calculus  Sarita, Saturday, April 3, 2010 at 5:52pm
thank you!
Additionally,
C) They give a recursively defined sequence: a_1=0.3; a_(n+1)=sqrt((a_n)+1)for n>1
How do you find out the first five terms for it. then prove that this sequence converges. What is a specific theorem that will guarantee convergence, along with the algebraic results of parts A and B?
* calculus  Damon, Saturday, April 3, 2010 at 7:26pm
.3
sqrt 1.3 = 1.14
sqrt 2.14 = 1.46
sqrt 2.46 = 1.57
sqrt 2.57 = 1.60
hmmm, not getting bigger very fast.
let's see what happens to the derivative for large n
.5/sqrt(x+1)
ah ha, look at that. When n gets big, the derivative goes to zero. So the function stops changing.
* calculus  Sarita, Sunday, April 4, 2010 at 1:05pm
But why would you look for the derivative to go to zero? Does it have to do anything with the theorem: If summation of a_n converges then limit_(n>infinity) of a_n = 0. If so, what would the limit be approaching? 10 or infinity? But if not, then what theorem would we use? I know you explained about the larger n for the derivative, but I do not understand how that relates to one of the theorems.
* calculus  Sarita, Sunday, April 4, 2010 at 1:21pm
But doesn't it converge to infinity and not 0?
* calculus  Sarita, Sunday, April 4, 2010 at 1:22pm
we want it to converge to 0 right? But does it even converge if it goes to infinity, or is that divergence?
* calculus  Sarita, Sunday, April 4, 2010 at 2:28pm
Do you do the limit on the derivative?
Or is there another way to prove convergence with a theorem of some sort?

calculus 
Damon,
Look at the sequence of 6 that we did. The number is getting closer and closer to 1.6 about. We are not looking at the sum of a series. We are looking at terms in a sequence. If they do not change for large n, they converged. They do not have to converge to zero. They just have to stop changing. Only their change (the derivative) must be zero for convergence.

calculus 
Sarita,
Which theorem, together with the results of parts a and b, will guarantee convergence?
Would it be the convergent sequences are bounded theorem, where if {a_n} converges, then {a_n} is bounded,
or
would it be the bounded monotonic sequences converge theorem, where 1) if {a_n} is increasing and a_n(</=)M for all n, then {a_n} converges and lim_(n>infinity)a_n(</=)M, and 2) if {a_n} is decreasing and a_n(>/=)m for all n, then {a_n} converges and lim_(n>infinity)a_n(>/=)m?
or
is there another one? 
calculus 
Damon,
Sure, that will do.

calculus 
Sarita,
So, what would be the exact limit of the sequence defined in part c? But it says to square the recursive equation and take limits using some limit theorems. How do you do that?