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    I have this question, which i seem to be able to do the most part of, but can't seem to figure out how to complete it..

    Poove, by induction, [1, 1, 0, 1] ^x = [1, x, 0, 1] for all values of x.

    Where the matrix is a 2x2 and reads across such that [A, B, C, D]

    thanks =)
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    For your induction hypothesis assume [1, 1, 0, 1] ^k = [1, k, 0, 1] for some value of k.

    Consider [1, 1, 0, 1] ^(k+1) = something X something...

    Can you take it from there?
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    ive gotten that far...it's atually going through the inductive part that i get stuck with...i can't seem to actually proove it. But i can tell that im close..there is something that im just not doing right, and i can't work out what...

    but thank you
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    I was doing this at college today! you say 'let it be true for x = k, and then write it the same but with k instead of x. Then you say, show to be true for x = k +1; which when you write it, is the matrice you wrote with k (matrice ^k = [1 k 0 1]) *times* the original matrice Because: something to the power of ^k+1 is the same as something to the power of k *times* soemthing to the power of one (just the original number/matrice) (rules of indices) - then you just multiply them and should be able to factor k+1 where x was at the start, hence showing it's true for k and K+1 therefore all values of x

    hope at least some part of that made any sense...!
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    (Original post by macccj)
    I was doing this at college today! you say 'let it be true for x = k, and then write it the same but with k instead of x. Then you say, show to be true for x = k +1; which when you write it, is the matrice you wrote with k (matrice ^k = [1 k 0 1]) *times* the original matrice Because: something to the power of ^k+1 is the same as something to the power of k *times* soemthing to the power of one (just the original number/matrice) (rules of indices) - then you just multiply them and should be able to factor k+1 where x was at the start, hence showing it's true for k and K+1 therefore all values of x

    hope at least some part of that made any sense...!
    yeah that makes loads of sense...thank you for that!! although, i might be being stupid...but then dont i get: [1, k, 0, 1] + [1, k+1, 0, 1] which doesnt get me the right answer?

    unless i've confused myself and added something extra?
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    how come you're adding? if you end up with [ 1, k+1, 0, 1] after multiplying your matrices', you've hence shown that it's true... i think :L
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    hold up - you multiply [1, 1, 0, 1]^k by [ 1, 1, 0, 1]

    which means you multiply [ 1, k, 0, 1] (because you said at the start let it be true that [1, 1, 0, 1]^k = [1, k, 0,1]) by [1, 1, 0, 1]
    which should work out as [1, k+1, 0, 1]?
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    (Original post by macccj)
    hold up - you multiply [1, 1, 0, 1]^k by [ 1, 1, 0, 1]

    which means you multiply [ 1, k, 0, 1] (because you said at the start let it be true that [1, 1, 0, 1]^k = [1, k, 0,1]) by [1, 1, 0, 1]
    which should work out as [1, k+1, 0, 1]?
    yeah, you're right....that makes sense now... I had confused myslef by looking at the wrong thing when i was working it out jsut now...Thank you for your help! It is much appreciated!
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    no worries, glad you got it!
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    i have another one for you today...proof by induction again..

    3^n > n^3 when n >3 ???

    am i right by showing it's true for n=4 as that is greater than 3?

    and then writing: assume true for n=k such that 3^k > k^3

    i've then gone on to say that the next term is 3^k+1 therefore,: 3^k + 3^k+1 > k^3 + 3^k+1 ??

    thanks
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    Yes you are right by showing it for n=4.

    You need to show that 3^{k+1} > (k+1)^3, given the assumption that 3^{k} > k^3. What you've written doesn't show that.
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    (Original post by Dragon)
    Yes you are right by showing it for n=4.

    You need to show that 3^{k+1} > (k+1)^3, given the assumption that 3^{k} > k^3. What you've written doesn't show that.

    thats what i was thinking..its what confused me..so how do i go about showing that 3^{k+1} > (k+1)^3 im guessing where i was going wrong was thinking that i added them ut i dont need to do i...i just have to prove 3^{k+1} > (k+1)^3 ?

    thank you for helping
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    Expanding both sides is a good start.
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    yeah, when you multiply it out you get multiples of the given formula from the start, sorry i was no more help!!
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    thank you =)
 
 
 
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