Prove Progression Proof of v(n+1)-3

  • Thread starter Andrax
  • Start date
  • Tags
    Proof
In summary, the problem involves finding the value of v(n+1)-3 and proving that it is equal to 1/2(v(n)-3). The approach involves using induction and manipulating the squared-up form of the equation.
  • #1
Andrax
117
0

Homework Statement


we have u(1)=1
u(n+1)=[itex]\frac{1}{16}[/itex](1+4u(n)+[itex]\sqrt{1+24u(n)}[/itex])

v(n)^2=1+24u(n)

Prove that [itex]\foralln[/itex][itex]\inN*[/itex]: v(n+1)-3=[itex]\frac{1}{2}[/itex](v(n)-3)

Homework Equations


The Attempt at a Solution


by working on v(n+1)-3 and replacing it by [itex]\sqrt{1+24u(n+1)}[/itex] didn't get me anywhere just a bunch of calc
also i made 6u(n+1)=[itex]\frac{1}{16}[/itex](v(n)+[itex]\frac{1}{2}[/itex])^2+[itex]\frac{19}{64}[/itex] but it didn't work for me , also just to mention the progression is not arithmetic or anything it's just (u(n)):n[itex]\geq1[/itex]
PS: the "^2" means squared
 
Last edited:
Physics news on Phys.org
  • #2
Have you tried induction? Btw, there's a parenthesis missing in un+1=(1/16)(1+4un+√(1+24un). Should it be un+1=(1/16)(1+4un)+√(1+24un)?
 
  • #3
haruspex said:
Have you tried induction? Btw, there's a parenthesis missing in un+1=(1/16)(1+4un+√(1+24un). Should it be un+1=(1/16)(1+4un)+√(1+24un)?

sorry forgot the parenthesis , using induction will make things worse ethe problem is the square root it's causing the problem here..
 
  • #4
Induction doesn't look too bad. Don't get tangled up with square roots - work with the squared-up form. Assume true up to
vn-3=(vn-1-3)/2
i.e.
vn=(vn-1+3)/2
Now work with vn+12 and see if you can show it equal to ((vn+3)/2)2. (Work from both ends.)
 
  • #5
Thanks solved it.
 

1. What is a progression proof?

A progression proof is a mathematical technique used to show that a given statement or formula is true for all natural numbers by proving its truth for a base case and then showing that if it holds for any arbitrary case, it must also hold for the next case. This technique is commonly used in mathematical induction.

2. What does v(n+1)-3 mean in a progression proof?

In a progression proof, v(n+1)-3 refers to the next case or step in the progression. It represents the value of the statement or formula being proven for the number n+1, which is the next natural number after n. Adding or subtracting a constant value, such as 3 in this case, is a common way to demonstrate progression in mathematical proofs.

3. How can one prove the progression proof of v(n+1)-3?

The progression proof of v(n+1)-3 can be proven using mathematical induction. First, the statement or formula must be shown to be true for a base case, typically n=1 or n=0. Then, it must be demonstrated that if the statement is true for any arbitrary case, such as n=k, then it must also hold for the next case, n=k+1. This process is repeated until it can be shown that the statement is true for all natural numbers.

4. What is the importance of proving progression proofs?

Proving progression proofs is important in mathematics because it allows us to establish the truth of a statement or formula for all natural numbers, rather than just a finite set of values. This is especially useful in fields like number theory and algebra, where proving a statement for all natural numbers is necessary to establish its validity.

5. Are there any common mistakes to avoid when proving progression proofs?

Yes, one common mistake to avoid when proving progression proofs is assuming that the statement is true for all natural numbers without actually proving it. Another mistake is using the wrong base case, which can lead to incorrect conclusions about the truth of the statement. It is also important to clearly explain the progression between cases and avoid making logical leaps in the proof.

Similar threads

  • Calculus and Beyond Homework Help
Replies
13
Views
627
  • Calculus and Beyond Homework Help
Replies
1
Views
522
  • Calculus and Beyond Homework Help
Replies
1
Views
434
  • Calculus and Beyond Homework Help
Replies
6
Views
398
  • Calculus and Beyond Homework Help
Replies
7
Views
302
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
482
  • Calculus and Beyond Homework Help
Replies
9
Views
846
  • Calculus and Beyond Homework Help
Replies
20
Views
1K
  • Calculus and Beyond Homework Help
Replies
24
Views
617
Back
Top