The Physics of Euler's Formula | Laplace Transform Prelude

0:00 [Submit subtitle corrections at criblate.com] This is the first

0:02 video in a trilogy aimed at demystifying the Laplace transform,

0:04 a powerful tool for studying differential equations.

0:08 Although we won't dig into the Laplace

0:10 transform itself until the next two chapters,

0:12 everything that we cover here sets up the mental frameworks and the prerequisite

0:17 knowledge that make understanding that transform as easy as I know how.

0:21 This video is a lot more than just preamble, though.

0:24 It's a very fun lesson in its own

0:26 right about how one of the most famous equations

0:28 in all of math is enables a bizarre trick

0:31 for solving an equation that is used ubiquitously throughout physics.

0:35 The main characters throughout this chapter

0:38 and the next two are exponential functions,

0:40 and I'm always going to be writing these as e^(st).

0:44 Here we think of t as being time and then s

0:47 as just some number determining which specific exponential we're talking about.

0:52 One of the big aims of this video is to motivate using physics why

0:56 it's useful to give s the freedom to take on not only real number values,

1:00 but complex ones as well.

1:02 But wait a minute, what does it even

1:04 mean to shove a complex number into an exponent?

1:07 Here I imagine there's a bit of a divide in the audience.

1:10 You see, there's some regular viewers

1:12 of math videos online for whom the specific

1:14 case of plugging inπ times i is a little bit cliched by this point.

1:18 It is amply covered by many videos on YouTube, but on the other hand,

1:23 most students find this to be an understandably baffling notion.

1:27 Given how absolutely fundamental this is to everything that follows,

1:30 even if you fall into that first camp,

1:32 I hope you'll agree that it's worth kicking things off here by reviewing

1:36 a very beautiful and visual way to understand what this idea is all about.

1:40 The nice part is that what follows doubles as a gentle

1:43 warm up for visualizing and thinking about differential equations.

1:51 You start with the fact that e to the t is its own derivative.

1:55 And really, you should think of this as what defines the number e.

1:58 Exponentials with other bases will have

2:01 derivatives that are proportional to themselves,

2:03 but e is the special number such that that proportionality constant is 1.

2:07 Now, very often in a calculus class,

2:09 you visualize derivatives as slopes of graphs.

2:11 That's all well and good, but it's not the only way.

2:13 You should get in the habit of flexing your mind a bit more.

2:16 For example, let's say you think of this as telling you the position

2:19 of some point on the number line as a function of time.

2:22 Then what the derivative expression is telling you is that at every

2:26 moment the velocity vector must look identical to the position vector.

2:30 And more specifically, because you know that e^0 equals 1,

2:33 anything to the 0 is 1, you also have an initial condition.

2:37 It's telling you you start at the number 1.

2:39 So at the very first moment,

2:41 the velocity is also 1, meaning it's pointed to the right.

2:44 But the farther to the right the position gets, the faster it must move.

2:48 So even if you had never heard of the function e^t or exponential growth,

2:52 this property alone is enough to give you a very visceral feeling

2:56 for how it gives a value that grows and at an accelerating rate.

3:00 But what if there was some constant in that exponent like e^(2t)?

3:04 Well, by the chain rule, the derivative is then two times the function itself.

3:09 And reading this dynamically,

3:10 it's telling you the velocity vector is always two times the position vector.

3:15 Again, the farther to the right the position gets, the faster it must move.

3:18 But this time the feeling is that that growth

3:20 gets out of hand all the more quickly.

3:22 What if that constant was negative, say negative 0.5?

3:26 Well, once again, by the chain rule,

3:29 the derivative of this function is minus 0.5 times the function itself.

3:33 So at every moment in time,

3:35 that velocity vector looks like a 180 degree rotation of the position vector,

3:39 but scaled down to be half its length.

3:42 This means you start moving to the left,

3:44 but as you approach zero with a smaller position vector,

3:48 that velocity must get proportionally smaller.

3:50 So it approaches zero, but at an ever slowing pace.

3:54 This, of course, is exponential decay.

3:56 But now for the fun part, why we're here in the first place.

3:59 What if that constant was an imaginary number i,

4:02 the square root of negative one?

4:04 Again, the chain rule tells us that the derivative

4:06 of this function is going to be i times the function itself.

4:10 Geometrically multiplying by i acts like a 90 degree rotation,

4:14 so this is telling you velocity always has to be perpendicular to position.

4:19 For anyone who's a little bit rusty

4:20 or needs a quick review with complex numbers,

4:23 let's say you have an arbitrary complex value a+ bi,

4:25 which you typically draw in the 2D plane.

4:27 Like this, the easiest way to think

4:30 about multiplying by i is to go component-by-component.

4:33 a times i ends up on this vertical imaginary line.

4:36 And then bi times i is b(i^2), which is -b.

4:41 This is where you actually use the defining property of i.

4:44 You'll notice each of those two individual components is rotated 90 degrees.

4:48 So the sum as a whole also has to get rotated 90 degrees.

4:52 Looking back at the equation, where this time,

4:54 evidently we must be thinking about position as being in the complex plane.

4:58 Even if you had never heard of raising e to an imaginary number,

5:02 and even if it's not clear at first what that would actually mean,

5:06 the expression is telling you that this value has to move in such a way

5:10 that the velocity vector is always locked to be

5:12 a 90 degree rotated copy of the position vector.

5:15 The only motion that satisfies this criterion is rotation around a circle,

5:20 and you can be more specific because the initial position is 1,

5:24 that velocity vector always has a unit length.

5:26 So this tells you how fast you have to move.

5:29 Your point wanders around that circle in such a way that it

5:32 traces one unit of arc length for every unit of time.

5:35 For example, to get one of the most famous equations in all of math,

5:39 if you wait forπ units of time, you end up precisely halfway around the circle.

5:44 This is why e^(πi) is -1.

5:48 One thing that's always worth emphasizing to any students initially

5:51 confused by this expression is just how misleading the notation is.

5:54 When you input a complex value,

5:56 the expression really has very little to do with repeated multiplication,

6:00 and honestly, not that much to do with the number e.

6:04 The computation that it refers to is

6:06 plugging in the input into this infinite polynomial, the Taylor series for e^x.

6:10 It's actually very fun, I think,

6:12 to take a moment to interpret the literal meaning of plugging

6:15 in something likeπ times i for each one of these polynomial terms.

6:19 As you take a higher power,

6:21 each extra factor of i rotates you another 90 degrees,

6:24 and then the π^n/ (n!) terms initially grow,

6:27 but then they shrink as that denominator takes over,

6:30 and you end up with this spiraling sum that converges to minus one.

6:35 Now, that said, it is not at all obvious,

6:38 just looking at this infinite polynomial,

6:39 that if you change the value t in the expression e^(i*t),

6:42 you're going to end up walking around a unit circle.

6:45 This is why focusing on the property of being its own

6:48 derivative is a lot more helpful than focusing on the underlying computation.

6:52 In practice, people get used to using

6:54 the expression e^x as a notational shorthand here,

6:57 and really think nothing of it.

6:59 Throughout this lesson, I'm going to be pulling up a complex

7:03 plane representing possible values for this number s.

7:05 And for each point on this plane, I want you to be able

7:09 to think about the corresponding exponential function e^(st).

7:11 We were just talking through what happens when s is equal to i.

7:15 On the lower right I can show you how that output changes with time.

7:19 And then on the top right, I might show a graph where,

7:21 in order to fit it on the screen,

7:23 I'll typically only graph the real part of that output with respect to time.

7:27 In this case, it looks like a cosine wave.

7:29 If s is a different imaginary number, something like two i,

7:33 it means you rotate at a different rate.

7:35 And on the one hand, this is obvious.

7:37 Throwing in a 2 in front of the time obviously moves you twice as fast.

7:41 But again, I think it's kind of fun

7:42 to read what the derivative expression is telling you dynamically.

7:46 In this case, you can read it as saying the velocity

7:48 is always a copy of the position vector rotated 90 degrees,

7:52 but stretched to have a length of 2.

7:55 More generally, it's common to label

7:57 this imaginary part with the Greek letter omega,

7:59 which describes the angular frequency of the motion.

8:02 In other words, how many radians of arc length

8:05 does it traverse around the circle per unit time?

8:08 This is just the imaginary axis.

8:10 But what about when s has both a real and an imaginary part?

8:14 Something like minus 0.5 plus i.

8:17 On the one hand, you can split up the exponential,

8:20 and this part here is telling you that the magnitude decays over time.

8:23 And then this part is telling you that there's rotation.

8:26 That's all well and good, but for fun, another way that you can think about

8:30 it is to continue with the previous intuition.

8:32 To understand what multiplication by any complex number looks like,

8:36 you can ask what combination of rotation and stretching would

8:39 place the vector at 1 onto this new value s.

8:42 For this example here, that would look like rotating a little over

8:45 90 degrees and stretching it out a bit.

8:48 So the derivative expression is telling you this is

8:50 always the relationship between the position and velocity vectors.

8:54 And I think geometrically, this gives you a very visceral way

8:57 to see why the motion must be spiraling inwards.

9:00 You'll sometimes hear engineers refer to this as the S plane,

9:03 and which essentially means you should think of each

9:06 point on that plane as encoding the entire function e^(st).

9:10 The imaginary part of s is always telling you

9:12 how rapidly the function oscillates and in which direction.

9:16 And then the real part of s is

9:18 telling you whether the magnitude grows or shrinks.

9:20 Positive real parts corresponding to exponential growth,

9:23 negative real parts corresponding to exponential decay.

9:28 This key equation telling us that velocity is some modified version of position.

9:32 Is basically a differential equation.

9:35 And for you and me, having just seen how an intuition for reading

9:39 off a differential equation like this can explain what complex exponents mean,

9:43 at least if you want e^x to retain its core property.

9:47 Let's flip this around to get back

9:49 to our core question and see how a different set

9:51 of differential equations can motivate why you would

9:54 ever care about complex exponents in the first place.

9:57 The easiest case to show you what I mean

9:59 by this is a very central example used all throughout physics,

10:02 where you imagine having a mass on a spring.

10:05 We're going to describe the position of this mass as x,

10:08 which will change over time.

10:10 The value 0 is going to correspond to the equilibrium position.

10:13 The derivative of this position versus time function,

10:16 of course, gives you the velocity of that mass.

10:18 And then the second derivative,

10:19 the rate of change of the velocity, gives you its acceleration.

10:23 The key feature of this spring setup is

10:25 that the more you pull that spring to one side,

10:28 the more strongly it accelerates the mass towards that equilibrium position.

10:33 More specifically, we say that the force

10:35 which is equal to mass times acceleration,

10:38 that's Newton's second law, is often well approximated as k times the position.

10:42 K here is just some positive proportionality constant.

10:46 It tells you how strong the spring is.

10:47 And this whole equation is telling you force is proportional to position.

10:51 Now, the reason this example is used

10:53 all throughout physics is because there are lots

10:55 of other situations where you approximate a force

10:58 and being proportional to some kind of offset.

11:00 Often it's not exactly that, but as a first order approximation,

11:03 it really helps you model what's going to happen.

11:06 It's also common to include a term here proportional to the velocity.

11:09 We call it a damping term.

11:11 This Greek letter mu is representing another positive coefficient.

11:15 And the negative sign is telling you the faster this mass is moving,

11:19 the stronger that damping force.

11:21 Maybe you think of it as friction, maybe you think of it as air resistance.

11:24 I remember as a physics student, always being bothered by the fact that neither

11:27 friction nor air resistance actually behave like this.

11:29 But the better way to view it is that, again,

11:32 this is a first order approximation of whatever

11:34 the slowing forces might be on this mass.

11:38 The point is, we now have a differential equation.

11:40 The position over time is an unknown function,

11:43 but we know it has to adhere to this constraint.

11:45 And your physical intuition probably tells you loosely

11:48 what you expect the solution to look like here.

11:50 There's going to be some oscillation back and forth,

11:53 and then as it loses energy to whatever those damping effects are,

11:56 the amplitude of that oscillation is going to decay.

11:59 I'm going to go ahead and take this equation and move everything to one

12:02 side of it so that we're setting a bunch of stuff equal to zero.

12:05 As a quick reminder, with differential equations,

12:07 there's not one specific function that solves it, per se.

12:10 For different initial positions where this mass might be,

12:13 you're going to get distinct functions that also solve the equation.

12:18 And that's actually only one out of two free parameters that we can change here,

12:21 because all of these are solutions where the initial velocity is zero.

12:25 But you could also imagine that the mass

12:27 starts out with some other non-zero velocity,

12:29 and every one of these combinations of an initial position and initial

12:33 velocity corresponds to a distinct function that also solves the equation.

12:37 So to solve this, really you're looking for a family of functions that solve it.

12:41 And preferably, you'd like some way to be able to narrow down

12:44 which member of that family solves it for your specific initial conditions.

12:49 So how do you solve it?

12:50 Well, there's this one very bizarre trick which I remember really

12:53 bothered me when I was a calculus student who first saw it,

12:57 which is where you simply guess that the answer looks like e^(st),

13:00 where s is just some constant, something that you're going to solve for.

13:04 The reason this really bothered me is that guessing and checking like this sort

13:08 of just feels like asking the student to know the answer ahead of time.

13:11 And also your physical intuition is telling you that an exponential

13:14 is probably not really how this mass on a spring behaves.

13:17 And yes, this is, frankly, unsystematic.

13:20 But the point I want to make is

13:21 that a desire to make this trick more systematic and more

13:24 generalizable is going to be one of the things

13:26 that leads you and me to the Laplace transform.

13:28 Right here, let's just run forward and see what it gives us.

13:32 If that position versus time really did look like e^(st),

13:35 then when you take its derivative, you.

13:38 You get the same function, but by the chain rule, you multiply it by s,

13:41 and then the second derivative again looks like the same function,

13:45 but it's picked up another factor of s.

13:47 And then all of the other constants just kind of come along for the ride.

13:51 What's very nice here is that you can factor out that e^(st),

13:55 and now everything that depends on time is tied up in this term right here.

14:00 And moreover, exponentials will never equal zero.

14:04 So if this equation is going to be true,

14:06 it means that this part right here has to equal zero.

14:09 So what you're left with is a piece of algebra.

14:11 Solve this quadratic equation, one that looks kind of like a mirror

14:15 image of the original differential equation that we had.

14:18 The easiest case here is if we ignore that damping coefficient.

14:22 Basically setting mu equal to zero with a little

14:25 bit of rearrangement and taking a square root,

14:28 what you find is that s is going to be plus or minus the square root of -k/m.

14:32 Now, k and m are both positive numbers,

14:35 so that means, whether you wanted it or not,

14:37 i, the square root of negative one, has now entered the game.

14:42 This square root of k/m term is something

14:44 that I'm going to give the suggestive shorthand name omega.

14:47 And rolling back, remember what it is that s represents.

14:51 We were exploring the possibility that a solution

14:54 to this equation looks like e^(st).

14:56 If we plug in these values for s, you now know what that means.

15:00 Plugging in a purely imaginary term like

15:02 this corresponds to oscillation in the complex plane.

15:06 Now, on the one hand, that is very weird because obviously our mass

15:09 on a spring needs a real valued solution, not these complex functions.

15:13 But on the other hand,

15:15 the idea of oscillating kind of matches what you want to find.

15:18 And it matches it quantitatively too.

15:20 Imagine that you increase that value k, meaning you have a stronger spring.

15:24 Well, then omega goes up.

15:25 So that corresponds to faster oscillation.

15:27 And your physical intuition backs that up.

15:29 A stronger spring probably would give you faster oscillation.

15:34 Even still, the result, frankly, feels bizarre, if not obviously nonsense.

15:38 I mean, the position of the mass on a spring is clearly a real number.

15:42 And if you zoom out, really what's going on here is that we found

15:46 for the pure mathematical equation divorced from any physics,

15:50 there exists a complex valued function that solves it, namely e^(i* omega* t).

15:55 To connect this pure mathland answer to something that's actually physical,

15:59 you need to squeeze out a real valued solution from this.

16:03 And the animation on screen kind of gives

16:05 you one indication of how you could do this.

16:07 You could just ignore the imaginary part,

16:09 only consider the real component of this solution that does actually work.

16:13 But a better way to think about it,

16:15 which will line up with the overall story I want to tell here that navigates

16:20 towards Laplace transforms is to add up

16:22 the two distinct complex solutions that we just found.

16:25 When you add these rotating vectors tip to tail,

16:27 the result stays constrained to the real number line.

16:30 And in fact, the way that it oscillates on that number line over time

16:34 looks like the function two times the cosine

16:37 of that same frequency term times T.

16:39 Now, the reason that you're allowed to just add two different solutions like

16:43 this to get another solution is based on a critical property of our equation.

16:48 It's what we call a linear equation,

16:50 which means if you have two distinct functions that solve it,

16:53 then when you add up those functions,

16:55 that sum of the two functions also solves the differential equation.

17:00 And actually you have more flexibility than that.

17:02 If you scale each one of those functions by some constant and you add them up,

17:06 that scaled sum is also a solution of the equation.

17:10 Remember, in solving an equation like this, we're not

17:12 just looking for one function or even two functions.

17:15 We're looking for a family of a whole bunch

17:17 of solutions that will depend on the initial conditions.

17:21 In this case, when we tried our admittedly random looking guess,

17:24 and the math came back to us with two distinct functions because it's linear,

17:28 you can scale each one of those functions by some constant,

17:31 add them together, and get a valid solution to the equation.

17:34 And those scaling coefficients don't have to be real numbers.

17:37 Those could also be complex numbers,

17:39 which in this case affects the initial angle of each of those rotating vectors.

17:44 The family of all possible functions you can get by tuning these two

17:48 coefficients is the family of all possible solutions to the original equation.

17:52 And most of these solutions are complex valued functions.

17:55 But the real valued solutions are a special case of those.

17:58 And which one you want depends on the initial conditions.

18:01 For example, if the initial position is supposed to be

18:04 2 and the initial velocity is supposed to be 0,

18:06 then you get a valid answer by setting both of these coefficients to be 1,

18:10 basically meaning you're just adding the two solutions we found earlier.

18:14 If the initial position is something different,

18:16 then you simply scale both those constants by the same amount.

18:21 Now, as presented so far, if this is supposed to be an example

18:24 of why complex exponents are a natural and desirable thing,

18:27 one of you could rightfully complain.

18:30 This is all just needlessly complicated!

18:32 If the so-called strategy is to just guess some function with a free parameter,

18:36 it's not like it's hard to guess for this situation

18:38 that a cosine or a sine would solve the equation,

18:41 and you could have that frequency term

18:43 be the free parameter that you're solving for.

18:45 What you would find if you knew to make this guess,

18:48 is that either cosine or sine can totally solve this equation,

18:51 as long as you set that frequency to be the square root of k over m.

18:55 And then, just as before, because this is a linear equation,

18:58 you can get the full family of solutions

19:01 by scaling both of these and adding them together.

19:03 And this is another valid way to describe the family of solutions.

19:07 Essentially, we're describing it with an alternate coordinate system.

19:10 And you could argue this is a way more sensible

19:12 coordinate system to use when we care about real solutions.

19:15 Because in this case, all the real solutions are what you get simply

19:19 by setting those scaling coefficients to be real numbers.

19:22 Isn't this just way more sensible?

19:24 Why complicate things with complex numbers?

19:26 The value of putting exponentials front and center makes

19:29 itself clear as soon as we try to generalize things.

19:32 So far, when we solved for s, we got these two different values in the complex

19:37 plane that are constrained to the imaginary line.

19:40 And as you change what the constants k and m look like,

19:43 you end up with different imaginary values that for your solution,

19:46 correspond to distinct frequencies in the oscillation that you get.

19:50 But think about what it means if we reintroduce that damping coefficient mu,

19:54 setting it to something that's not equal to zero.

19:57 Well, in this case, solving the equation

19:59 looks like applying the quadratic formula.

20:01 And you don't really need to dwell on the details of the algebra here.

20:04 I'm just going to go ahead and show you what it

20:07 looks like if I increase the value of that coefficient mu,

20:10 and we see where the two corresponding

20:13 solutions for S land in the complex plane.

20:16 The salient feature is that they have not

20:18 only an imaginary but also a negative real component.

20:21 And just a few minutes ago, we talked all about what it looks like if you want

20:26 to exponentiate something with a negative real part and an imaginary part,

20:29 it both does decays and oscillates,

20:31 where the real part tells you how much it decays,

20:34 the imaginary part tells you how much it oscillates.

20:37 In this case, what I'll do is graph

20:38 for you the real component of that exponential,

20:40 and it kind of matches what you would expect of the spring.

20:43 One thing that's actually pretty fun here is

20:46 how if you increase that damping coefficient mu enough,

20:48 eventually the solutions no longer have any imaginary

20:51 part and they only have a real component,

20:53 meaning the solution just looks like decay.

20:56 And when this happens, you call the spring overdamped.

20:59 This whole example is called the damped harmonic oscillator.

21:02 Like I said, it's very fundamental throughout physics,

21:04 so just understanding it in its own right is a worthy enough task.

21:08 But how far does this dumb little trick actually take us?

21:12 The straightforward way that you can generalize it is for any equation

21:15 that looks like this, where you're taking a bunch of higher order derivatives,

21:19 you're scaling each one by some constant, you add them all up,

21:22 and you set the result equal to zero.

21:24 In that case, everything we just did works essentially the same way.

21:28 If you substitute e^(st) for x,

21:30 then all of these derivative terms look just like

21:33 that, but each one picks up an additional factor of s.

21:36 This lets you factor out all of the exponential parts,

21:39 leaving you with a certain polynomial in s that you want to equal zero.

21:44 One of the most fundamental facts in algebra,

21:47 literally called the fundamental theorem of algebra,

21:50 is that polynomials can always be factored into linear

21:53 terms like this, exposing n roots to the equation,

21:56 as long as you give those roots

21:59 the freedom to maybe take on complex number values.

22:02 So, for example, if this was some fifth degree equation,

22:05 your solutions might look something like this in the S-plane.

22:08 Just like the oscillator example, this is basically the math telling you, hey,

22:13 e^(st) can absolutely be a valid solution as long

22:15 as you set s equal to one of these values.

22:18 And just as before, this is a linear equation.

22:21 So you can find the family of all solutions

22:24 by scaling each one of these exponentials and adding them together.

22:27 All of these constants are like knobs and dials

22:30 that you can tune to your heart's content.

22:32 They can be real or they can be complex,

22:34 allowing them to influence both the amplitude and the phase of each term.

22:37 The specific values will depend on your initial conditions.

22:41 I am glossing over a certain nuance when it comes to repeated roots,

22:44 but this is the general idea.

22:46 Unfortunately, most real world equations are not simple linear ones like this.

22:51 For example, the equation for a damped harmonic oscillator has actually

22:54 come up on this channel before in a video about optics,

22:57 but it came with a twist.

22:59 We were studying why light appears to slow down in a medium like glass,

23:03 causing it to refract.

23:04 And the key question was to understand why

23:06 this depends on the color of that light, giving the effect of a prism.

23:10 Now, I'm not going to recount all the details here,

23:13 but what you need to know is that deep in that video,

23:16 we were modeling charges inside the material,

23:18 like glass, as little damped harmonic oscillators.

23:21 These little charges wiggling about some equilibrium position,

23:24 were being influenced by an external force,

23:26 in this case, an incoming light wave,

23:28 which oscillated up and down as a sine wave.

23:31 And critically, the frequency of that incoming light would in general

23:34 have nothing to do with the natural resonant frequency of the oscillator.

23:39 So, in short, we were studying the same equation,

23:41 but with this added term that looks like a certain cosine expression.

23:44 Now, unlike the linear case,

23:46 the family of solutions here does not look as simple as a linear combination

23:50 of exponentials where you can freely tune

23:52 all of those constants to your heart's content.

23:54 And this dumb trick of just guessing e^(st) certainly is not going to work.

23:59 However, everything that we've discussed here does actually

24:01 bring you a lot closer than you might expect.

24:04 The solutions in this more complicated case do happen

24:07 to look like a combination of four specific exponentials.

24:11 It's just that, unlike the linear case,

24:12 you can't freely tune all of the coefficients.

24:15 It's a lot more constrained.

24:16 In fact, the whole substance of that prism example comes down to understanding

24:21 exactly how big these coefficients are

24:23 as a function of that incoming light frequency.

24:26 This is a surprisingly common outcome

24:28 where the solutions to some differential equation

24:30 that pops up in the real world looks like a certain combination of exponentials,

24:34 but with particular coefficients.

24:36 This ubiquity of exponentials is why engineers benefit from an intuitive

24:41 understanding of points on the S-plane and how they can encode growth,

24:45 decay, and oscillation.

24:47 You can kind of think about these functions,

24:50 e^(st) as being like the atoms of calculus.

24:53 What I mean by that is that complicated functions that describe

24:56 our world can often be broken up into these parts.

24:58 And as long as you give s the freedom to take on complex values,

25:02 and by breaking it up that way, they become simpler to understand and to study.

25:06 This becomes especially true if you allow for infinite combinations,

25:10 potentially over a continuum of values for s rather than some discrete set.

25:15 We're going to go deep with that idea,

25:17 and it is hard to overstate how powerful it is.

25:19 The key question is,

25:21 given some unknown function and a differential equation describing it,

25:24 even if you assume it can be broken up into exponential

25:26 parts like this, how do you actually find what those parts are?

25:31 Taking the forced harmonic oscillator, for example:

25:33 How would you know that the solution is built out of four specific exponentials?

25:37 How would you solve for the appropriate values of s in the exponents?

25:41 And how would you solve for the corresponding

25:43 coefficients for a particular initial condition?

25:46 There is a tool for this job, and as you may have guessed by this point,

25:50 it's something known as a Laplace Transform.

25:52 If you watched the earlier chapter about Fourier series,

25:55 a lot of what I'm saying here is probably ringing all kinds of bells:

25:59 Imaginary exponentials as describing a kind of rotation breaking

26:03 up general functions as a sum of those rotating exponentials.

26:06 And there is absolutely a connection here.

26:08 A big part of the story I want to tell is how this Laplace

26:12 transform we are building up to extends

26:14 the notion of Fourier series and Fourier transforms,

26:16 applying to a much more general family of functions.

26:20 We're going to go into much more detail in the next two chapters,

26:23 but here's a high level preview.

26:24 When you use a Laplace transform to solve a differential equation,

26:28 it actually ends up looking remarkably similar

26:30 to that dumb trick of substituting e^(st).

26:34 In the context of our dumb trick,

26:36 the differential equation turned into algebra basically because the act

26:39 of taking a derivative is the same as multiplication by s,

26:43 at least for these specific functions.

26:46 That same thing happens when you use a Laplace transform,

26:49 and it's for essentially the same reason too.

26:52 What that operation does is translate functions

26:54 into a new language where these terms e^(st),

26:58 the atoms of calculus, are the fundamental units.

27:01 Then again, the fact that differentiation in time looks like multiplication

27:05 by s for these terms means that in this new language,

27:09 derivatives start to look a lot like multiplication,

27:12 and differential equations start to look like algebra.

27:16 To see how exactly this transform is defined,

27:18 how you can visualize what it's doing,

27:21 and how to use it to concretely solve a nonlinear equation,

27:24 come join me in the next chapter.

27:27 At the time I'm publishing this, an early

27:29 view for that next chapter is available on Patreon,

27:31 and my plan is to incorporate the feedback

27:34 and get a finalized version out by next week.

27:36 See you then.