00:01
In this problem we have fibonacci numbers and the golden ratio.
00:07
So this is quite a beautiful problem.
00:12
We have some sub questions so let's get started with the first one.
00:17
We are going to here list the first 15 fibonacci numbers.
00:24
So we start from one and the second one is again one and we just add the last two terms together so two three and two plus three five three plus five eight and so on we have 13 21 34 55 89 89 14 1445 89 144 24 24 24 24 23 777 and this should be the last one 610.
01:08
These are the first 15 fibonacci numbers.
01:15
Okay we are going to discuss this number five, namely the golden ratio.
01:23
It comes from the following.
01:26
Okay, first the slide is done.
01:29
Golden ratio.
01:34
We are going to state what this number, precise number is.
01:41
And what it means and then we are going to determine what sequence of numbers approaches this value five so let's get started with the definition if that makes sense or if that is the correct word for that so suppose you have some line here a straight line from between a and c then you take a point here b and the golden ratio is defined as follows.
02:17
We say that the ratio of the length of this entire piece, entire line from between a and c, to the length of the line from a to b.
02:34
This should be equal to the ratio of the line segment between a and b to the ratio of the line segment between b.
02:46
And see so that is how this number emerges in mathematical or in nature basically so okay let us label these lengths using fractions or some relative factors i'm going to say that that or assert that this length should be equal to 1 and i'm going to scale these other lengths accordingly so i will call this length phi which is greater than 1 using our equality then the entire length becomes phi plus 1 okay now let's work on this equality using phi and this other length 1 we have phi plus 1 over 5 equal to 5 over 1 so we have this qualitative equation 5 squared equal to 5 plus 1 or 5 squared minus 5 minus 1 equal to 0 and we know how to solve this 5 equal to 1 plus minus 1 plus 4 divide by 2 now there are 2 roots and as we can see one of them is less than 1 but we want the root that is larger than 1 so we take the positive root or the root with the plus sign so we have five equal to one plus square of five over two so this is the precise value or the exact value of this number and it goes like 1 .6113 so this is the definition of this number this goes ratio.
05:24
Now it comes from also from a sequence, some sequence of numbers, and this sequence was discovered by fibonacci.
05:35
It goes like this.
05:37
Suppose i define my sequence by a sub n.
05:42
And fibonacci discovered that if you take the ratio of two consecutive fibonacci numbers, you as as n goes to infinity you will get you'll get this golden ratio okay so let's elaborate on this but let me just indicate that this is the final answer so this is the final answer and where we have f sub n as the fibonacci numbers and and goes like one two three etc so let us write a couple of terms for this sequence again.
06:38
So let us remember some values for the fibon numbers from the first question.
06:43
We have one, one, two, three, five, eight, thirteen, twenty one, thirty, four, fifty five, eighty -nine.
07:02
144 so this should be enough this should be enough for us to observe the limit actually so again let us take the ratio of the first two terms so it is one over one then this ratio the third term to the second term which is two then three or 2 it is 1 .5 okay it will oscillate a bit at the beginning but it will converge to the golden ratio quite fast so we have next 5 over 3 it is 1 .66 6 7 then 8 over 5 1 .6 13 over 8 1 .6 1 .6 to 1 .6 to 5 .5 1 .6 to 5 and you get the idea 20 over 13 34 over 21 etc so let me just write down the rest of the terms so we are already in the neighborhood of this, the actual value of this number.
08:42
So this is the sequence namely the ratio of two consecutive fibibund numbers is the sequence that gives the golden ratio as n goes to basically infinity.
09:01
Okay, the third question.
09:05
We are going to do some tiny algebra two times some tiny arithmetics here one plus one over one plus one over one so we have one plus one plus one plus one one plus one over two so this is three over two okay the next one we have two plus two plus two plus two 2 plus 2 over 2 over 2.
09:43
So 2 plus 2 over 2 plus 1 2 plus 2 over 3.
09:56
So 2 plus 2 over 3.
10:07
So we have 3 plus 2 over 3 or 3 plus 3 over 3.
10:09
So we have 3 plus 3 or 3 plus 3 over 3.
10:10
So we have 3 plus 3 plus 3 or 3.
10:13
So we have 3 plus 3 plus 1.
10:15
1 is equal to 3 plus 3 over 4 so we have 15 over 4 these are the answers to this question next we have two equations to solve separately of course because we have only one unknown the first equation is x is equal to 2 plus 1 over x so let's equate the denominators by multiplying both sides by x...