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?
4 answers
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?