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?
===========================
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?
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?
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?
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.
Sure, that will do.
To prove that if 0 (≤) x (≤) 10, then 0 (≤) sqrt(x+1) (≤) 10, we can follow these steps:
Step 1: Check the increasing behavior of the square root function.
- Take the derivative of sqrt(x+1) with respect to x: d(sqrt(x+1))/dx = 0.5/sqrt(x+1).
- Notice that the derivative is positive for all x in the domain (i.e., 0 to 10).
- This means that the square root function is increasing throughout the domain.
Step 2: Evaluate the endpoints of the domain.
- When x = 0, sqrt(x+1) = sqrt(1) = 1.
- When x = 10, sqrt(x+1) = sqrt(11) ≈ 3.32.
- Therefore, the inequality holds: 1 (≤) sqrt(x+1) (≤) 3.32 for 0 (≤) x (≤) 10.
To prove that if 0 (≤) u (≤) v (≤) 10, then 0 (≤) sqrt(u+1) (≤) sqrt(v+1) (≤) 10, we can use the same approach:
Step 1: Check the increasing behavior of the square root function.
- The derivative of sqrt(x+1) is still positive for all x in the domain (0 to 10).
- Therefore, the square root function is increasing throughout the domain.
Step 2: Apply the logic of the previous proof to each inequality in the sequence.
- If u is less than or equal to v, then sqrt(u+1) is less than or equal to sqrt(v+1).
- If sqrt(u+1) is less than or equal to sqrt(v+1), then sqrt(v+1) is less than or equal to 10.
- Therefore, the compound inequality 0 (≤) sqrt(u+1) (≤) sqrt(v+1) (≤) 10 holds for 0 (≤) u (≤) v (≤) 10.
Now, for the recursive sequence given by a_1 = 0.3 and a_(n+1) = sqrt(a_n +1) for n > 1:
To find the first five terms of the sequence, apply the recursive definition:
- a_1 = 0.3
- a_2 = sqrt(a_1 + 1) = sqrt(0.3 + 1) ≈ 1.14
- a_3 = sqrt(a_2 + 1) = sqrt(1.14 + 1) ≈ 1.46
- a_4 = sqrt(a_3 + 1) ≈ 1.57
- a_5 = sqrt(a_4 + 1) ≈ 1.60
To prove that this sequence converges, we observe that for larger values of n, the derivative of sqrt(x+1) approaches 0. This indicates that the function stops changing as n gets large, which suggests convergence. This result aligns with the specific theorem that states if a sequence has a limit as n approaches infinity, then the derivative of the sequence tends to 0. By checking the derivative, we can be assured that the sequence converges.
Another theorem that guarantees convergence is the Monotone Convergence Theorem (also known as the Bolzano-Weierstrass Theorem). This theorem states that if a sequence is bounded (i.e., it remains within a set range) and monotonic (either non-decreasing or non-increasing), then it must converge.
In this case, we have shown that the sequence is bounded between 0 and 10 (as per the previous proofs) and that it increases (as per the increasing nature of the square root function). Therefore, by applying the Monotone Convergence Theorem, we can conclude that the sequence converges.