載入中...

相關課程

登入觀看

⇐ Use this menu to view and help create subtitles for this video in many different languages.
You'll probably want to hide YouTube's captions if using these subtitles.

相關課程

0 / 750

- In the last video, I showed the Laplace transform of t, or
- we could view that as t the first power, is equal to 1/s
- squared, if we assume that s is greater than 0.
- In this video, we're going to see if we can generalize this
- by trying to figure out the Laplace transform of t to the
- n, where n is any integer power greater than 0, so n is
- any positive integer greater than 0.
- So let's try it out.
- So we know from our definition of the Laplace transform that
- the Laplace transform of t to the n is equal to the integral
- from 0 to infinity of our function-- well, let me write
- t to the n-- times, and this is just the definition of the
- transform, e to the minus st, dt.
- And similar to when we figured out this Laplace transform,
- your intuition might be that, hey, we should use integration
- by parts, and I showed it in the last video.
- I always forget it, but I just recorded it, so I do happen to
- remember it.
- So the integration by parts just tells us that the
- integral of uv prime is equal to uv minus the integral of--
- I view this as kind of the swap-- so u prime v.
- So this is just our integration by parts formula.
- If you ever forget it, you can derive it in about 30 seconds
- from the product rule.
- And I did it in the last video because I hadn't used it for
- awhile, so I had to rederive it.
- So let's apply it here.
- So what do we want to make our v prime?
- It's always good to use the exponential function, because
- that's easy to take the antiderivative of.
- So this is our v prime, in which case our v is just the
- antiderivative of that.
- So it's e to the minus st over minus s.
- If we take the derivative of this, minus s divided by minus
- s cancels out, and you just get that.
- And then if we make our u-- let me pick a good color here.
- If we make this equal to our u, what's our u prime?
- u prime is just going to be n times t to the n minus 1.
- Fair enough.
- So let's apply the integration by parts.
- So this is going to be equal to uv.
- u, I'll use this t to the n, so u is t to the n, that's our
- u, times v, which is e-- let me write this down-- so it's
- minus-- there's a minus sign there, so we put the minus.
- Let me do it in that color.
- I'm just rewriting this. e to the minus st/s.
- So that's the uv term right there.
- Let me make that clear.
- And let me pick a good color here.
- So this term right here is this term right here.
- And, of course, I'm going to have to evaluate this from 0
- to infinity, so let me write that: 0 to infinity.
- I could put a little bracket there or something, but you
- know we're going to have to evaluate that.
- And then from that, we're going to have to
- subtract the integral.
- And let me not forget our boundaries.
- 0 to infinity of u prime is n times t to the n minus 1--
- that's our u prime-- times v times minus-- so let me put
- this minus out here-- so minus e to the minus st/s.
- And then all of that.
- Of course, we have our dt, and you have a minus minus.
- These things become pluses.
- Let's see if we can simplify this a little bit.
- So we get our Laplace transform of t to the n is
- equal to this evaluated at infinity and evaluated at 0.
- So when you've evaluate-- what's the limit of this as t
- approaches infinity?
- As t approaches infinity, this term, you might say, oh, this
- becomes really big.
- And I went over this in the last video.
- But this term overpowers it, because you're going to have e
- to the minus infinity, if we assume that s is
- greater than 0.
- So if s is greater than 0, this term is going to win out
- and go to 0 much faster than this term is
- going to go to infinity.
- So when you evaluate it at infinity, when you evaluate
- this at infinity, you're going to get 0.
- And then you're going to subtract this evaluated at 0.
- This evaluated at 0, when it's evaluated at 0 is just minus 0
- to the n times e to the minus s times 0/s Well, this
- becomes 0 as well.
- So this whole term evaluated from 0 to infinity is all 0,
- which is a nice, convenient thing for us.
- And then we're going to have this next term right there.
- So let's take out the constant terms. This n
- and this s are constant.
- They are constant with respect to t.
- So you have plus n/s times the integral from 0 to infinity of
- t to the n minus 1 times e to the minus st, dt.
- Now, this should look reasonably familiar to you.
- What's the definition of the Laplace transform?
- The Laplace transform of any function is equal to the
- integral from 0 to infinity of that function times e to the
- minus st, dt.
- Well, when we have an e to the minus st, dt, we're taking the
- integral from 0 to infinity, so this whole integral is
- equal to the Laplace transform of this, of t
- to the n minus 1.
- So just that easily, because this term went to 0, we've
- simplified things.
- We get the Laplace transform of t to the n is equal to--
- this is all 0-- it's equal to n/s-- that's right there--
- times this integral right here, which we just figured
- out was the Laplace transform of t to the n minus 1.
- Well, this is a nice, neat simplification.
- We can now figure out the Laplace transform of a higher
- power in terms of the one power lower that, but it still
- doesn't give me a generalized formula.
- So let's see if we can use this with this information to
- get a generalized formula.
- So the Laplace transform of just t-- so let me write that
- down; I wrote that at the beginning of the problem.
- We get the Laplace transform, I could write this as t to the
- 1, which is just t, is equal to 1/s squared, where s is
- greater than 0.
- Now, what happens if we take the Laplace
- transform of t squared?
- Well, we can just use this formula up here.
- The Laplace transform of t squared is equal to 2/s times
- the Laplace transform of t, of just t to the 1, right?
- 2 minus 1.
- So times the Laplace transform of t to the 1.
- Well, t, we know what that is.
- This is equal to 2/s times this, times 1/s squared, which
- is equal to 2/s to the third.
- Interesting.
- Let's see if we can do another one.
- What is-- I'll do it in the dark blue-- the Laplace
- transform of t to the third?
- Well, we just use this formula up here.
- It's n/s.
- In this case, n is 3.
- So it's 3/s times the Laplace transform of t to the n minus
- 1, so t squared.
- We know what the Laplace transform of this one was.
- This is just this right there.
- So it's equal to 3/s times this thing.
- And I'm going to actually write it this way, because I
- think it's interesting.
- So I'll write the numerator.
- Times 2 times 1/s over s squared, which is-- we could
- write it as 3 factorial over-- what is this? s
- to the fourth power.
- Let's do another one.
- And I think you already are getting the idea of
- what's going on.
- The Laplace transform of t to the fourth power is what?
- It's equal to 4/s times the Laplace transform of t the
- third power.
- And that's just 4/s times this.
- So it's 4/s times 3 factorial over s to the fourth.
- So now 4 times 3 factorial, that's just 4 factorial over s
- to the fifth.
- And so you can just get the general principle.
- And we can prove this by induction.
- It's almost trivial based on what we've already done.
- That the Laplace transform of t to the n is equal to n
- factorial over s to the n plus 1.
- We proved it directly for this base case right here.
- This is 1 factorial over s to the 1 plus 1.
- And then if we know it's true for this, we know it's going
- to be true for the next increment.
- So induction proof is almost obvious, but you can even see
- it based on this.
- If you have to figure out the Laplace transform of t to the
- tenth, you could just keep doing this over and over
- again, but I think you see the pattern pretty clearly.
- So anyway, I thought that was a neat problem in and of
- itself, outside of the fact, it'll be useful when we figure
- out inverse and Laplace transforms. But this is a
- pretty neat result.
- The Laplace transform of t to the n, where n is some integer
- greater than 0 is equal to n factorial over s to the n plus
- 1, where s is also greater than 0.
- That was an assumption we had to make early on when we took
- our limits as t approaches infinity.
- Anyway, hopefully you found that useful.