Working out help (induction) (1 Viewer)

HeroicPandas · Aug 16, 2013

Hi, could anyone please check my working out?

Question: $$If $u_1 = 1$ and u_n = \sqrt{3u_{n-1}}$ $ for n \geq 2,$ $ show that u_n <3$ $ for $n\geq 1$ and deduce that u_{n+1} > u_n$ $ for n\geq 1$

$1. u_n = \sqrt{3u_{n-1}},$ RTP: u_n <3$ \\ $Show true for $n = 2, u_2 = \sqrt{3u_1} = \sqrt{3} <3 \\ $Assume true for $n = k, u_k<3 \\ $Prove true for $n = k+1, \\ i.e. u_{k+1} = \sqrt{3u_k} \\ ~~~~~~~~~~~<\sqrt{3\cdot 3} ($using assumption) \\ $~~~~~~~~~~~= 3 \\ $\therefore $ $ By mathematical induction $u_n <3$ is true for all positive integers (pretend I typed out the conclusion for induction) \\ \\ 2. u_{k+1}>u_k \\ \\ LHS = u_{k+1} \\ ~~~~~~~~= \sqrt{3u_k} = \sqrt{3\sqrt{3u_{k-1}}} = \sqrt{3\sqrt{3\sqrt{3\sqrt{3\dots}}}} \\ \\ $Let \alpha = \sqrt{3\sqrt{3\sqrt{3\sqrt{3\dots}}}} \\ \therefore \alpha(\alpha-3) = 0, $ $ since \alpha >0, \therefore \alpha = 3 \\ \\ ~~~~~LHS = 3 \\ \\$

$\therefore $ This results in $3 > u_n,$ which is true from part (i)\\\\ Is this solution valid?$

Drongoski · Aug 16, 2013

HeroicPandas said:
Hi, could any please check my working out?

Question: $$If $u_1 = 1$ and u_n = \sqrt{3u_{n-1}}$ $ for n \geq 2,$ $ show that u_n <2$ $ for $n\geq 1$ and deduce that u_{n+1} > u_n$ $ for n\geq 1$

u_n < 2 or < 3 ??

HeroicPandas · Aug 16, 2013

Drongoski said:
u_n < 2 or < 3 ??

oh sorry, u_n <3, thanks for noticing that!

Carrotsticks · Aug 16, 2013

First part is correct, though you didn't really need to 'pretend' to have written the conclusion. It was already done right there.

The last part isn't valid. The reason is because you're assuming that the nested roots are infinitely nested in order to have $\alpha ^2=3\alpha$

Carrotsticks · Aug 16, 2013

If you want a hint for the second part, try considering $\frac{u_{k+1}}{u_k}$

seanieg89 · Aug 16, 2013

No, stay away from infinitely nested things in the HSC unless they are given to you! This is just an induction question.

$u_{n+1}=\sqrt{3u_n}>\sqrt{u_n^2}=u_n$

from the earlier result that $u_n<3$ , as well as the fact that from the definition it is clear that each $u_n$ is positive.

HeroicPandas · Aug 16, 2013

Carrotsticks said:
First part is correct, though you didn't really need to 'pretend' to have written the conclusion. It was already done right there.

The last part isn't valid. The reason is because you're assuming that the nested roots are infinitely nested in order to have $\alpha ^2=3\alpha$

ok i understand thanks

seanieg89 said:
No, stay away from infinitely nested things in the HSC unless they are given to you! This is just an induction question.

$u_{n+1}=\sqrt{3u_n}>\sqrt{u_n^2}=u_n$

from the earlier result that $u_n<3$ , as well as the fact that from the definition it is clear that each $u_n$ is positive.

ahh i cant believe i could not see that.. thanks

seanieg89 · Aug 16, 2013

No worries, glad to help

.

imblue · Aug 16, 2013

how do i post my own question?

HeroicPandas · Aug 16, 2013

Carrotsticks said:
If you want a hint for the second part, try considering $\frac{u_{k+1}}{u_k}$

nvm

Working out help (induction) (1 Viewer)

HeroicPandas

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Drongoski

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HeroicPandas

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Carrotsticks

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seanieg89

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HeroicPandas

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imblue

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HeroicPandas

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