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Thread: Finding x1000

  1. #1
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    Post Finding x1000

    Hi there, I am completely stuck and would like some help with this. Id appreciate it very much:

    Given that x1=5 and xD=2xn-1 -5 for n>1. Find x1000

    Finding x1000-d42b8409-7fc3-4b69-946a-8a9ffd1526ef.jpeg

    Thanks again!
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  2. #2
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    Re: Finding x1000

    Quote Originally Posted by Asiram View Post
    Hi there, I am completely stuck and would like some help with this. Id appreciate it very much:

    Given that x1=5 and xD=2xn-1 -5 for n>1. Find x1000

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    $(\forall n)[x_n=5]$ see here.
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    Re: Finding x1000

    I think I understand where that came from. How would I use this to find x1000?

    Thanks!
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  4. #4
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    Re: Finding x1000

    Quote Originally Posted by Plato View Post
    $(\forall n)[x_n=5]$ see here.
    Did you not understand what this meant? $\displaystyle x_{1000}= 5$!
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  5. #5
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    Re: Finding x1000

    Quote Originally Posted by Asiram View Post
    Hi there, I am completely stuck and would like some help with this. I’d appreciate it very much:

    Given that x1=5 and xD=2xn-1 -5 for n>1. Find x10

    Thanks again!
    You mean "xn", not "xD". The first thing I would do is calculate a few terms. [tex]x_1= 5[/tex[ so $\displaystyle x_2= 2x_1- 5= 2(5)- 5= 10- 5= 5$. Then $\displaystyle x_3= 2x_2- 5= 2(5)- 5= 5$. Then $\displaystyle x_4= 2x_3- 5= 2(5)- 5= 10- 5= 5$. See a pattern? Let's make the "guess" that, for all n, $\displaystyle x_n= 5$ and prove it by "induction".

    We are given that $\displaystyle x_1= 5$. Suppose that, for some k, $\displaystyle x_k= 5$. Then [tex]x_{k+1}= 2x_k- 5= 2(5)- 5= 10- 5= 5! By "induction", [tex]x_n= 5[/rex] for all n. "Induction", here, allows us to extend the fact that, at each step we are doing the same thing, to all positive integers.
    Thanks from Archie
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    Re: Finding x1000

    Quote Originally Posted by HallsofIvy View Post
    You mean "xn", not "xD". The first thing I would do is calculate a few terms. [tex]x_1= 5[/tex[ so $\displaystyle x_2= 2x_1- 5= 2(5)- 5= 10- 5= 5$. Then $\displaystyle x_3= 2x_2- 5= 2(5)- 5= 5$. Then $\displaystyle x_4= 2x_3- 5= 2(5)- 5= 10- 5= 5$. See a pattern? Let's make the "guess" that, for all n, $\displaystyle x_n= 5$ and prove it by "induction".

    We are given that $\displaystyle x_1= 5$. Suppose that, for some k, $\displaystyle x_k= 5$. Then [tex]x_{k+1}= 2x_k- 5= 2(5)- 5= 10- 5= 5! By "induction", [tex]x_n= 5[/rex] for all n. "Induction", here, allows us to extend the fact that, at each step we are doing the same thing, to all positive integers.
    Yeah, that was a typo. Thank you both for your help. Truly appreciate it!
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  7. #7
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    Re: Finding x1000

    Quote Originally Posted by Asiram View Post
    Given that x1=5 and xD=2xn-1 -5 for n>1. Find x1000
    Still get a headache reading that "mess"...

    How would it be worded if "properly worded"?
    x1 = 5 means 1st term of an arithmetic sequence?
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    Re: Finding x1000

    Quote Originally Posted by DenisB View Post
    How would it be worded if "properly worded"?
    x1 = 5 means 1st term of an arithmetic sequence?
    First of all it is clearly not an arithmetic sequence.
    It is a recursively defined sequence: the first term is $x_1=5$, then for $n>1,~x_n=2\cdot x_{n-1}-5$.
    That is each term after first term, each tern is twice the preceding term minus five.
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  9. #9
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    Re: Finding x1000

    Thanks for the Tylenol Plato!
    I'll send you a 12ouncer of Canadian Club for Xmas...
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