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14:26
The other way to visualize derivatives | Chapter 12, Essence of calculus
3Blue1Brown
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May 12, 2026
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Transcript
0:07
The months ahead of you hold within them a lot of hard work, some neat examples,
0:11
some not-so-neat examples, beautiful connections to physics,
0:14
not-so-beautiful piles of formulas to memorize,
0:17
plenty of moments of getting stuck and banging your head into a wall,
0:20
a few nice aha moments sprinkled in as well, and some genuinely lovely
0:24
graphical intuition to help guide you through it all.
0:27
But if the course ahead of you is anything like my first introduction to calculus,
0:31
or any of the first courses I've seen in the years since,
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0:34
there's one topic you will not see, but which I believe stands to greatly accelerate
0:38
your learning.
0:40
You see, almost all of the visual intuitions from that first year are based on graphs.
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The derivative is the slope of a graph, the integral is a certain area under that graph.
0:50
But as you generalize calculus beyond functions whose inputs and outputs are
0:54
simply numbers, it's not always possible to graph the function you're analyzing.
1:00
So if all your intuitions for the fundamental ideas, like derivatives,
1:04
are rooted too rigidly in graphs, it can make for a very tall and largely
1:08
unnecessary conceptual hurdle between you and the more quote-unquote advanced
1:13
topics like multivariable calculus and complex analysis, differential geometry.
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1:18
What I want to share with you is a way to think about derivatives,
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which I'll refer to as the transformational view,
1:24
that generalizes more seamlessly into some of those more general contexts
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where calculus comes up.
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And then we'll use this alternate view to analyze a fun puzzle about repeated fractions.
1:35
But first off, I just want to make sure we're all
1:37
on the same page about what the standard visual is.
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If you were to graph a function, which simply takes real numbers as inputs and outputs,
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one of the first things you learn in a calculus course is that the derivative gives
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you the slope of this graph, where what we mean by that is that the derivative of
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the function is a new function which for every input x returns that slope.
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Now I'd encourage you not to think of this derivative
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as slope idea as being the definition of a derivative.
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Instead think of it as being more fundamentally about how
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sensitive the function is to tiny little nudges around the input.
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And the slope is just one way to think about that sensitivity
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relevant only to this particular way of viewing functions.
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I have not just another video, but a full series on this
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topic if it's something you want to learn more about.
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The basic idea behind the alternate visual for the derivative is to
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think of this function as mapping all of the input points on the
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number line to their corresponding outputs on a different number line.
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In this context, what the derivative gives you is a measure of how
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much the input space gets stretched or squished in various regions.
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That is, if you were to zoom in around a specific input and take a look at some
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evenly spaced points around it, the derivative of the function of that input is
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going to tell you how spread out or contracted those points become after the mapping.
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Here, a specific example helps.
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Take the function x2, it maps 1 to 1, 2 to 4, 3 to 9, and so on.
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You can also see how it acts on all of the points in between.
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If you were to zoom in on a little cluster of points around the input 1,
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and see where they land around the relevant output,
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which for this function also happens to be 1, you'd notice that they tend
3:23
to get stretched out.
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In fact, it roughly looks like stretching out by a factor of 2.
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The closer you zoom in, the more this local behavior looks just like multiplying by a
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factor of 2. This is what it means for the derivative of x2 at the input x equals 1 to be
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2.
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It's what that fact looks like in the context of transformations.
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If you looked at a neighborhood of points around the input 3,
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they would get stretched out by a factor of 6.
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This is what it means for the derivative of this function at the input 3 to equal 6.
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Around the input 1 fourth, a small region tends to get contracted specifically by a
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factor of 1 half, and that's what it looks like for a derivative to be smaller than 1.
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The input 0 is interesting.
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Zooming in by a factor of 10, it doesn't really
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look like a constant stretching or squishing.
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For one thing, all of the outputs end up on the right positive side of things.
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As you zoom in closer and closer, by 100x, or by 1000x,
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it looks more and more like a small neighborhood of points around 0 just
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gets collapsed into 0 itself. This is what it looks like for the derivative to be 0.
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The local behavior looks more and more like multiplying the whole number line by 0.
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It doesn't have to completely collapse everything to a point at a particular zoom level,
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instead it's a matter of what the limiting behavior is as you zoom in closer and closer.
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It's also instructive to take a look at the negative inputs here.
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Things start to feel a little cramped since they collide with where all the positive
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input values go, and this is one of the downsides of thinking of functions as
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transformations.
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But for derivatives, we only really care about the local behavior anyway,
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what happens in a small range around a given input.
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Here, notice that the inputs in a little neighborhood around, say,
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negative 2, don't just get stretched out, they also get flipped around.
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Specifically, the action on such a neighborhood looks more
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and more like multiplying by negative 4 the closer you zoom in.
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This is what it looks like for the derivative of a function to be negative.
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And I think you get the point, this is all well and good,
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but let's see how this is actually useful in solving a problem.
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A friend of mine recently asked me a pretty fun question about the infinite
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fraction 1 plus 1 divided by 1 plus 1 divided by 1 plus 1 divided by 1,
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and clearly you watch math videos online, so maybe you've seen this before,
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but my friend's question actually cuts to something you might not have
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thought about before, relevant to the view of derivatives we're looking at here.
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The typical way you might evaluate an expression like this is to set it equal to x,
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and then notice that there is a copy of the full fraction inside itself.
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So you can replace that copy with another x, and then just solve for x.
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That is, what you want is to find a fixed point of the function 1 plus 1 divided by x.
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But here's the thing, there are actually two solutions for x,
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two special numbers where 1 plus 1 divided by that number gives you back the same thing.
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One is the golden ratio, phi, around 1.618, and the other is negative 0.618,
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which happens to be negative 1 divided by phi.
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I like to call this other number phi's little brother,
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since just about any property that phi has, this number also has.
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And this raises the question, would it be valid to say that the infinite
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fraction we saw is somehow also equal to phi's little brother, negative 0.618?
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Maybe you initially say, obviously not, everything on the left hand side is positive,
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so how could it possibly equal a negative number?
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Well, first we should be clear about what we actually mean by an expression like this.
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One way you could think about it, and it's not the only way,
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there's freedom for choice here, is to imagine starting with some constant, like 1,
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and then repeatedly applying the function 1 plus 1 divided by x, and then asking,
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what is this approach as you keep going?
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I mean, certainly symbolically what you get looks more and more
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like our infinite fraction, so maybe if you wanted to equal a number,
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you should ask what this series of numbers approaches.
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And if that's your view of things, maybe you start off with a negative number,
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so it's not so crazy for the whole expression to end up negative.
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After all, if you start with negative 1 divided by phi,
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then applying this function 1 plus 1 over x, you get back the same number,
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negative 1 divided by phi, so no matter how many times you apply it,
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you're staying fixed at this value.
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But even then, there is one reason you should
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view phi as the favorite brother in this pair.
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Here, try this, pull up a calculator of some kind, then start with any random number,
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and plug it into this function, 1 plus 1 divided by x,
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and plug that number into 1 plus 1 over x, and again, and again, and again, and again.
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No matter what constant you start with, you eventually end up at 1.618.
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Even if you start with a negative number, even one that's really close to phi's
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little brother, eventually it shies away from that value and jumps back over to phi.
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So, what's going on here?
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Why is one of these fixed points favored above the other one?
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Maybe you can already see how the transformational understanding of derivatives
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is helpful for understanding this setup, but for the sake of having a point of contrast,
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I want to show you how a problem like this is often taught using graphs.
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If you were to plug in some random input to this function,
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the y value tells you the corresponding output, right?
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So to think about plugging that output back into the function,
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you might first move horizontally until you hit the line y equals x,
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and that's going to give you a position where the x value corresponds to your
9:26
previous y value, right?
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So then from there, you can move vertically to see what output this new x value has,
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and then you repeat.
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You move horizontally to the line y equals x to find a point whose x value is the same
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as the output you just got, and then you move vertically to apply the function again.
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Now personally, I think this is kind of an awkward way
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to think about repeatedly applying a function, don't you?
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I mean, it makes sense, but you kind of have to pause
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and think about it to remember which way to draw the lines.
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And you can, if you want, think through what conditions make this spiderweb
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process narrow in on a fixed point, versus propagating away from it.
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In fact, go ahead, pause right now, and try to think it through as an exercise.
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It has to do with slopes.
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Or if you want to skip the exercise for something that I think gives a much more
10:15
satisfying understanding, think about how this function acts as a transformation.
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So I'm going to go ahead and start here by drawing a bunch of
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arrows to indicate where the various sampled input points will go.
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And side note, don't you think this gives a neat emergent pattern?
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I wasn't expecting this, but it was cool to see it pop up when animating.
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I guess the action of 1 divided by x gives this nice emergent circle,
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and then we're just shifting things over by 1.
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Anyway, I want you to think about what it means to repeatedly apply some function,
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like 1 plus 1 over x, in this context.
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Well after letting it map all of the inputs to the outputs,
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you could consider those as the new inputs, and then just apply the same process again,
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and then again, and do it however many times you want.
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Notice, in animating this with a few dots representing the sample points,
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it doesn't take many iterations at all before all of those dots kind of clump in around 1.
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618.
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Now remember, we know that 1.618 and its little brother,
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negative 0.618 on and on, stay fixed in place during each iteration of this process.
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But zoom in on a neighborhood around phi.
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During the map, points in that region get contracted around phi,
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meaning that the function 1 plus 1 over x has a derivative with a magnitude less
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than 1 at this input.
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In fact, this derivative works out to be around negative 0.38.
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So what that means is that each repeated application scrunches the neighborhood
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around this number smaller and smaller, like a gravitational pull towards phi.
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So now tell me what you think happens in the neighborhood of phi's little brother.
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Over there, the derivative actually has a magnitude larger than 1,
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so points near the fixed point are repelled away from it.
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And when you work it out, you can see that they get
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stretched by more than a factor of 2 in each iteration.
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They also get flipped around, because the derivative is negative here,
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but the salient fact for the sake of stability is just the magnitude.
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Mathematicians would call this right value a stable fixed point,
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and the left one is an unstable fixed point.
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Something is considered stable if when you perturb it just a little bit,
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it tends to come back towards where it started, rather than going away from it.
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So what we're seeing is a very useful little fact,
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that the stability of a fixed point is determined by whether or not the magnitude of its
12:45
derivative is bigger or smaller than 1.
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This explains why phi always shows up in the numerical play,
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where you're just hitting enter on your calculator over and over,
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but phi's little brother never does.
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As to whether or not you want to consider phi's little brother a
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valid value of the infinite fraction, well that's really up to you.
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Everything we just showed suggests that if you think of this expression
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as representing a limiting process, then because every possible seed
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value other than phi's little brother gives you a series converging to phi,
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it does feel silly to put them on equal footing with each other.
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But maybe you don't think of it as a limit, maybe the kind of math
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you're doing lends itself to treating this as a purely algebraic object,
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like the solutions of a polynomial, which simply has multiple values.
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Anyway, that's beside the point, and my point here is not that viewing derivatives
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as this change in density is somehow better than the graphical intuition on the whole.
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In fact, picturing an entire function this way can be
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kind of clunky and impractical as compared to graphs.
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My point is that it deserves more of a mention in most of the
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introductory calculus courses, because it can help make a
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student's understanding of the derivative a little more flexible.
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Like I mentioned, the real reason I'd recommend you carry this perspective
13:58
with you as you learn new topics is not so much for what it does with your
14:01
understanding of single variable calculus, it's for what comes after.
— end of transcript —
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