Koch snowflake fractal
Area of Koch Snowflake (part 2) - Advanced Summing an infinite geometric series to finally find the finite area of a Koch Snowflake
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- In the last video we got as far as figuring out that the area of this Koch snowflake
- This thing that has an infinite perimeter, can be expressed as this infinite sum over here
- So our job in this video is to try to simplify this, and hopefully get a finite value
- Let's do our best to actually simplify this thing right over here
- So the easiest part of this thing to simplify is this right over here
- So let's just focus on that
- Then if we can get a value for this part that I am bracketing off
- Then we can just place that value here and simplify the rest of it
- So what I've just bracketed off can be re-written as three times four ninths
- plus four ninths squared
- plus four ninths to the third power
- And you can go on and on and on
- Plus four ninths to every other power, all the way through infinity!
- Lucky for us, there is a way to figure out this infinite (geometric) series
- There's a way to figure this out, and I've done several videos where we prove the general thing
- But I'll just do it by hand this time, just so that we don't have to resort to some magical formulas
- So let's say that we define some sum, this one over here (let's call it S)
- Let's say that S is equal to what we have in parentheses over here
- It's going to be equal to four ninths, plus four ninths squared, plus four ninths to the third
- all the way to infinity
- Now let's also say that we multiply S by four ninths
- What's four ninths S going to look like?
- So then, I'm just essentially multiplying every term here by four ninths
- So if I take this first term and multiply it by four ninths, what am I going to get?
- Well I'm going to get four ninths squared
- If I take the second term and multiply it by four ninths, I'm going to get four ninths to the third power
- And we are going to go all the way to infinity
- So this is interesting
- When I multiply four ninths times this I get all of the terms here except for this first four ninths
- Now, this is kind of the magic of how we can actually find the sum of an infinite geometric series
- We can subtract this term right over here (this pink line) from this green line
- If we do that, clearly this is equal to that and this is equal to that
- So if we subtract this from that its equivalent to subtracting the pink from the green
- So we get S minus four ninths S is equal to...
- Well, every other term, this guy minus this guy is going to cancel out
- And that's going to happen all the way to infinity
- and on the right hand side you're only going to be left with this four ninths over here
- Then this four ninths, we can (S is the same thing as nine over nine) write this as nine over nine S
- minus four ninths S is equal to four ninths
- So nine over nine minus four over nine of something gives us five over nine
- So this becomes five ninths S is equal to four ninths
- Then to solve for S (and this is kind of magical but it's actually quite logical)
- Multiply both sides times the inverse of this, so times nine fifths on both sides
- These guys cancel out, and we get S is equal to four fifths
- That's really neat!
- We've just shown that this whole thing over here is equal to four fifths
- So this entire bracket that we did over here is equal to three times four fifths
- This entire bracket is equal to twelve over five
- Now let's go to our original expression, so we don't lose track of what we are doing
- We have this square root of three times S squared over sixteen
- Then we have this four here plus the entire thing in brackets (which simplified to twelve fifths)
- Just to add these two together we can rewrite four as twenty fifths
- and then twenty over five plus twelve over five is thirty-two over five
- Let me write that down over here
- This is the home stretch now, this is very exciting!
- We are about to find the finite area of something that has an infinite perimeter!
- So it's going to be square root of three times S squared over sixteen
- Times thirty-two over five
- We can divide the thirty-two in the numerator by the sixteen there, which gives two
- We are left with the area of a Koch snowflake
- (where the initial equalateral triangle that we started with has each of its sides as length S)
- is, two times the square root of three times S squared all over five
- For example, if that first equalateral triangle had a side-length of one
- Then the area of this crazy thing (that has an infinite perimeter) would just be two square roots of three over five
Be specific, and indicate a time in the video:
At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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