Monday

October 20, 2014

October 20, 2014

Posted by **Sarita** on Saturday, April 3, 2010 at 1:15am.

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?

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?

D) How do you find out the exact limit of the sequence defined in part C? Are you supposed to square the recursive equation and take limits using limit theorems? If so, then which are these theorems?

Thank you very much.

- calculus -
**Damon**, Saturday, April 3, 2010 at 8:43pmI answered this below after I did the first two.

- calculus -
**Damon**, Saturday, April 3, 2010 at 8:47pmcalculus - 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.

**Answer this Question**

**Related Questions**

Calculus - A) How do you prove that if 0(<or=)x(<or=)10, then 0(<or=)...

calculus - A) How do you prove that if 0(<or=)x(<or=)10, then 0(<or=)...

calculus - A) How do you prove that if 0(<or=)x(<or=)10, then 0(<or=)...

calculus - A) How do you prove that if 0(<or=)x(<or=)10, then 0(<or=)...

Math Help please!! - Could someone show me how to solve these problems step by ...

Calculus - Please look at my work below: Solve the initial-value problem. y'' + ...

calculus - A) How do you prove that if 0(<or=)x(<or=)10, then 0(<or=)...

calculus - prove that d/dx 4x .√(x + √x) = 6x+5 (x)1/2/√(x...

math calculus please help! - l = lim as x approaches 0 of x/(the square root of...

Math/Calculus - Solve the initial-value problem. Am I using the wrong value for ...