Author Topic: Fun questions :) (Read 114922 times) Tweet Share

TrueTears · « **Reply #150 on:** May 06, 2009, 04:25:26 pm »

CORRECT KAMIL! :coolsmiley:

/0 · « **Reply #151 on:** May 06, 2009, 05:52:57 pm »

well done

kamil9876 · « **Reply #152 on:** May 06, 2009, 06:00:37 pm »

this was for me a very awkward solution. first time i ever applied the geometric sequence to complex numbers, let alone the limit of it. I wasn't sure whether it would work, e.g: try to apply that formula to the sequence:

1-1+1-1+1....

First term is 1, ratio is -1. Formula gives 1/2

You may say this was incorrect since the formula is valid for for ratios r, such that -1<r<1. But in what sense does a complex number satisfy such an inequality? Fortunately i was able to justify the reasoning in my use of the geometric sequence in this case.

Something to think about:

what is the sum: 1+(1+i)+(1+i)^2...
how about 1+0.5(1+i)+(0.5(1+i))^2...

/0 · « **Reply #153 on:** May 06, 2009, 07:17:16 pm »

With real numbers we have |r|<1, with complex numbers it's |r|<1 in the sense of magnitude!

$\left|\frac{e^{i \theta}}{2}\right|=\frac{1}{2}<1$

I think it's amazing how it extends like that

kamil9876 · « **Reply #154 on:** May 06, 2009, 07:33:40 pm »

yep that's exactly how i justified that. Discovereed something new today

Never thought using geometric sequence in complex no.

kamil9876 · « **Reply #155 on:** May 06, 2009, 08:28:27 pm »

and here is a proof of that property:

$S=1+z+z^2...z^n$
$Sz=z+z^2...z^n$

$Sz-S=z^n-1$
$S=\frac{z^n-1}{z-1}$

Hence the formula is the same.

Now if $|z|<1$ then $\lim_{n\rightarrow \infty} |z^n|=0$ .

$0 \leq |Re(z^n)| \leq |z^n|$
$0 \leq |Im(z^n)| \leq |z^n|$

Hence sandwhich theorem gives:
$<br />\lim_{n\rightarrow \infty} |Re(z^n)|=0$
$\lim_{n\rightarrow \infty} |Im(z^n)|=0$

and finally it can be shown using this that:
$<br />\lim_{n\rightarrow \infty} S=\frac{1}{1-z}$

TrueTears · « **Reply #156 on:** May 07, 2009, 06:09:49 pm »

Nice kamil

TrueTears · « **Reply #157 on:** May 13, 2009, 10:15:45 pm »

43. Consider a rectangle l units long, and w units wide and the lw unit squares found in it. A diagonal is drawn from one corner to the opposite. The diagonal cuts through certain unit squares. Find in terms of l and w the number of unit squares that are cut through.
(sauce: kamil9876)

$l + w - GCD(l,w)$

kamil9876 · « **Reply #158 on:** May 13, 2009, 10:16:39 pm »

Quote from: TrueTears on May 13, 2009, 10:15:45 pm

43. Consider a rectangle l units long, and w units wide and the lw unit squares found in it. A diagonal is drawn from one corner to the opposite. The diagonal cuts through certain unit squares. Find in terms of l and w the number of unit squares that are cut through.
(sauce: kamil9876)

$l + w - GCD(l,w)$

correct, now the proof please ;p

kamil9876 · « **Reply #159 on:** May 25, 2009, 08:20:41 pm »

Quote

4. The Fibonacci ratios are defined as follows: $\frac{1}{1},\frac{2}{1},\frac{3}{2},\frac{5}{3}$ (i.e $r_n=\frac{f_{n+1}}{f_{n}}$ )
prove that $r_1<r_2>r_3<r_4...$ etc. (increase and decrease alternatively)
(source: kamil9876)

Here is a a verification for the first 999 ratios

Note that the program does not assume this is true, but merely compares the top to the bottom value.

2.0
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1.61764705882
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TrueTears · « **Reply #160 on:** May 25, 2009, 08:25:20 pm »

holy... that's amazing.

kamil9876 · « **Reply #161 on:** May 25, 2009, 08:30:43 pm »

Hhaha i just noticed that when it prints the number, it truncates to some fixed number of decimal places, but the computer still remembers the untruncated value.

To understand what im on about, look at the 75 at the end.

kamil9876 · « **Reply #162 on:** May 26, 2009, 11:18:21 pm »

Quote

69. Find $\frac{b}{a}$ if $\log_{9}a=\log_{15}b=\log_{25}(a+2b)$
(sauce: /0)

$log_{9}a=x$
$log_{15}b=x$
$log_{25}(a+2b)=x<br />$
Looking at first two equations:

$9^x=a$
$15^x=b$

Now I summon the assitance of my beloved prime factors :smitten::

$3^x 3^x=a$
$5^x 3^x=b$

$\frac{b}{a}=(\frac{5}{3})^x$

Now let's analyse the second and third equation because they have a common prime factor of 5 (omg i swear i love you guys):

$3^x 5^x=b$
$5^x 5^x=a+2b$

$(\frac{5}{3})^x=\frac{a}{b}+2$

Now we can equate:
$<br />\frac{b}{a}=\frac{a}{b}+2$
$let \frac{b}{a}=c$

$c=\frac{1}{c}+2$
$c^2=1+2c$
$c=1+\sqrt{2}$ (dropping the other solution as that would imply c being negative and that would mean a OR b must be negative, which we do cannot have)

TrueTears · « **Reply #163 on:** May 26, 2009, 11:18:52 pm »

correct!

lol gogo prime factors!

TrueTears · « **Reply #164 on:** June 19, 2009, 10:26:48 pm »

Added new questions by /0.

gogo!~~~

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Author Topic: Fun questions :) (Read 114922 times) Tweet Share

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Re: Fun questions :)

kamil9876

Re: Fun questions :)

kamil9876

Re: Fun questions :)

TrueTears

Re: Fun questions :)

kamil9876

Re: Fun questions :)

kamil9876

Re: Fun questions :)

TrueTears

Re: Fun questions :)

TrueTears

Re: Fun questions :)

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