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, Sunday, April 4, 2010 at 6:47pm
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, Monday, April 5, 2010 at 12:10am
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, Monday, April 5, 2010 at 6:27am
Sure, that will do.
* calculus - Sarita, Tuesday, April 6, 2010 at 4:05am
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?
To prove that if 0(≤)x(≤)10, then 0(≤)sqrt(x+1)(≤)10, we can use calculus and the concept of derivatives.
First, we need to determine if the square root function is increasing in the given domain (0 to 10). To do this, we can find the derivative of the function sqrt(x+1) with respect to x.
d(sqrt(x+1))/dx = 0.5/sqrt(x+1)
The derivative is positive for all values in the domain (0 to 10), which means that the square root function is increasing in that range.
Now, we can check the ends of the domain to prove the inequality.
If x = 0, then sqrt(x+1) = sqrt(1) = 1. So, 0(≤)sqrt(0+1)(≤)10 holds.
If x = 10, then sqrt(x+1) = sqrt(11) ≈ 3.32. So, 0(≤)sqrt(10+1)(≤)10 holds.
Therefore, we have proven that if 0(≤)x(≤)10, then 0(≤)sqrt(x+1)(≤)10.
For part B, we can use the same reasoning. Since the derivative of the square root function is positive throughout the domain, if v is greater than u, then sqrt(1+v) will be greater than sqrt(1+u). Thus, the inequality holds: 0(≤)sqrt(u+1)(≤)sqrt(v+1)(≤)10.
Moving on to part C, let's find the first five terms of the recursively defined sequence:
a1 = 0.3
a2 = sqrt(a1 + 1) = sqrt(0.3 + 1) ≈ 1.14
a3 = sqrt(a2 + 1) = sqrt(1.14 + 1) ≈ 1.46
a4 = sqrt(a3 + 1) = sqrt(1.46 + 1) ≈ 1.57
a5 = sqrt(a4 + 1) = sqrt(1.57 + 1) ≈ 1.60
To prove that this sequence converges, we can observe that as n (the term number) gets larger, the derivative of the function sqrt(x+1) with respect to x approaches zero. This means that the function stops changing significantly as n increases, indicating convergence.
The theorem that guarantees this convergence is the Bounded Monotonic Sequences Converge theorem, which states that if a sequence is increasing and bounded above, or decreasing and bounded below, then the sequence converges.
In this case, the sequence is increasing (as observed from the first five terms) and bounded above by 10 (as proved in parts A and B), so it converges. The exact limit of the sequence can be estimated to be around 1.60 based on the calculated terms.
Therefore, the sequence defined in part C converges.