Mathematical Induction

  • #1
AerospaceEng
28
0

Homework Statement



Consider the sequence of real numbers x1, x2, x3,.... defi ned by the relations x1 = 1

and xn+1 =[tex]\sqrt{1 + 2xn}[/tex]

1. Use mathematical induction to show that xn+1 > xn
for all n [tex]\geq[/tex] 1.

The Attempt at a Solution



I'm a bit thrown off by this question because it seems very obvious that it would be greater. if x1=1 is it safe to assume that xn=n? and if so I feel like this proof is stupid cause it shows right in front of you that its greater. any help is great thanks.
 

Answers and Replies

  • #2
36,203
8,192
Fixed your LaTeX.

Homework Statement



Consider the sequence of real numbers x1, x2, x3,.... defi ned by the relations x1 = 1

and xn+1 =[tex]\sqrt{1 + 2x_n}[/tex]

1. Use mathematical induction to show that xn+1 > xn
for all n [tex]\geq[/tex] 1.

The Attempt at a Solution



I'm a bit thrown off by this question because it seems very obvious that it would be greater. if x1=1 is it safe to assume that xn=n? and if so I feel like this proof is stupid cause it shows right in front of you that its greater. any help is great thanks.
The obviousness (or not) is irrelevant to what you need to do, which is to prove this statement by induction.

You are given x1 = 1. What is the value for x2? If x2 > x1, then use that for your base case.

Next, assume that xk > xk - 1 (the induction hypothesis), and use that assumption to show that xk + 1 > xk. If you can do this, you will have proved that the statement is true for all n >= 1.
 
  • #3
╔(σ_σ)╝
831
2
The proof is not stupid! Lol engineers!

Perform the substraction [tex] x_{n+1}- x_n[/tex] then show the substraction is > 0. Hint: rationalise and use the fact that a^2 >0 for an real number a.
 
  • #4
AerospaceEng
28
0
I'm I allowed to say that xn+1 is the same thing as xn + x1 cause i feel I can't and if I can't I feel so limited.
 
  • #5
Char. Limit
Gold Member
1,208
22
I'm I allowed to say that xn+1 is the same thing as xn + x1 cause i feel I can't and if I can't I feel so limited.

But that's not true.

xn+1 is clearly shown to be [tex]\sqrt{1+2x_n}[/tex].

So...

[tex]x_{n+1}-x_n = \sqrt{1+2x_n} - x_n[/tex]

So you need to prove that the above equation is always greater than 0, given x1=1.
 
  • #6
AerospaceEng
28
0
okay so I think i see where this is going. what i've done so far is i've taken the RHS and multiplied it by itself just with a positive between the terms, to give me 1+2xk-xk2

But i feel like i've done something wrong because i haven't used x1=1 as of yet..
 
Last edited:
  • #7
Char. Limit
Gold Member
1,208
22
okay so I think i see where this is going. what i've done so far is i've taken the LHS and multiplied it by itself just with a positive between the terms, to give me 1+2xk+xk2
and i can factor this to give me (x+1)2
Is that good enough to show that it's always greater than 0. and then ill just plug this mini proof into my main proof.

But i feel like i've done something wrong because i haven't used x1=1 as of yet..

Considering that x_k itself is always greater than 0, that should be enough. But I'll wait for a second opinion on that.
 
  • #8
AerospaceEng
28
0
now if i work with the left hand side I get xk+12-xk2 and i can cancel the -xk2 on both sides to give me:

xk+12=2k+1

and i feel im stuck again and where does the x1=1 come into play?
 
  • #9
hunt_mat
Homework Helper
1,760
27
the x_1 comes into play when you show that x_2>x_1 which is the initial step when you should have done first.
 
  • #10
36,203
8,192
now if i work with the left hand side I get xk+12-xk2 and i can cancel the -xk2 on both sides to give me:

xk+12=2k+1

and i feel im stuck again and where does the x1=1 come into play?
See post #2.
 

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