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Mathematical Induction |
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| Jan23-11, 12:53 PM | #1 |
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Mathematical Induction
1. The problem statement, all variables and given/known data
Consider the sequence of real numbers x1, x2, x3,.... defined 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. 3. 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. |
| Jan23-11, 01:50 PM | #2 |
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Mentor
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Fixed your LaTeX.
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. |
| Jan23-11, 02:40 PM | #3 |
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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. |
| Jan24-11, 03:51 AM | #4 |
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Mathematical Induction
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.
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| Jan24-11, 03:59 AM | #5 |
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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. |
| Jan24-11, 04:15 AM | #6 |
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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.. |
| Jan24-11, 04:20 AM | #7 |
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| Jan24-11, 04:38 AM | #8 |
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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? |
| Jan24-11, 06:16 AM | #9 |
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Recognitions:
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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.
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| Jan24-11, 09:31 AM | #10 |
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