# Kinematic equation looks like a Taylor series

1. May 19, 2010

### Jack21222

I was just pondering today how the kinematic equation for position looks like a taylor expansion.

x = x0 + dx/dt *t + (1/2)*d2x/dt2*t2

I believe it continues like that, exactly like a taylor expansion does, so the next term would be (1/6)*d3x/dt3*t3

If it is indeed a taylor expansion, what function is it a taylor expansion of? Just the function describing its motion? I feel like I should be able to figure this out on my own, but I'm just checking to make sure I'm not crazy.

2. May 19, 2010

### Acut

No, its not a Taylor Series expansion (despite the similarity). It does not approximate any function - it is the exact equation which describes a accelerated (constant acceleration) motion. Just integrate twice d2x/dt2=a

However, when the acceleration is not constant, but varies linearly with time, (i.e. d3x/dt3=b) you get an expression for x wich has the same shape of the equation you suggested (x = x0 + dx/dt *t + (1/2)*d2x/dt2*t2 + (1/6)*d3x/dt3*t3). Just integrate three times the above equation.

Yes, your intuition is quite correct!

3. May 20, 2010

No? It approximates an arbitrary position function doesnt it?

4. May 20, 2010

### Acut

Of course not. It may be an physical approximation (like any other law of Physics), but it's not a mathematical approximation (like Taylor series).
It's the exact equation, not an approximation.
As I've said before, just integrate d2x/dt2= a twice and you will get that equation.

5. May 20, 2010

I dont buy it. Its only an exact equation if you define the higher order terms to be zero.

6. May 21, 2010

### dst

This is correct. Of course, the higher order terms vanish very fast so we don't really need to consider them.

And yes, it is a taylor expansion of sorts except a Taylor expansion involves summing from 0 => infinity. It follows because in order to get the position, you need x(t=0), then add on the contributions from velocity by integrating the velocity wrt t then the acceleration is integrated twice, jerk (d^3 x/dt ) is integrated three times and so on and so forth. Supposing you had an infinite number of derivatives then this would become the Taylor expansion.

Last edited: May 21, 2010
7. May 21, 2010

### xxChrisxx

There are no higher order terms to more accutarely descrivbe it though. Each of those terms describes something different. If you increase the order of a taylor series is gives you more accurate information about the same thing.

It's the difference between me walking is a squiggly line (that would require a taylor series to approximate the path I took). But If I wanted to find my velocity and ecceleration you'd just differentiate/integrate as required.

Going to a higher order tells you nothing really that useful. Rate of change of acceleration (jerk), then rate of change of rate of change of acceleration.

EDIT: Also just as an asside the kinematic equations are constant acceleration. Meaning that the jerk is zero anyway.

Last edited: May 21, 2010
8. May 21, 2010

In any case, the kinematic equation is the first three terms of the taylor expansion of an arbitrary position function.

You can assume the higher order terms are negligible for the purposes of class or your application, you can define them to be zero for a canned textbook problem where acceleration is constant. Thats why we like the the taylor expansion, we can often safely ignore the higher order terms.

9. May 21, 2010

### Feldoh

Just to clarify you are right, well to an extent. Taylor series has the great property that just about any smooth continuous function can be expanded as a taylor series.

However as other people have stated we ordinarily do not look at the higher order terms when it comes to motion since the contribution continually gets smaller and smaller. In the equation you stated: $$x = x0 + dx/dt *t + (1/2)*d^2x/dt^2*t^2$$ we have a constant acceleration which (obviously) kills any higher terms so while you could still represent the position with a taylor series you'd wind up getting:

$$x = x0 + dx/dt *t + (1/2)*d^2x/dt^2*t^2+0+0+0+0+0+0+0+0+0+0+0+...$$ (you get the picture)

10. May 21, 2010

### arunma

Actually I think he's right. The equation of motion is only valid for situations in which the acceleration is constant. There are no higher order terms; they're zero by definition of "constant acceleration." As Jack has astutely observed, the equation of motion does look exactly like a Taylor series. But it isn't a Taylor series. You can actually derive it without even using calculus.

11. May 21, 2010

### Jack21222

Then, I meant "equations describing motion," I guess. I didn't realize calling it "kinematic" defined it as having constant acceleration. It looks like you guys understood what I meant anyway.

12. May 22, 2010

I dont generally think of them that specific either. To me, kinematic equations in general describe any type of classical motion - with or without a jerk (or snap, crackle and pop :) )

13. May 22, 2010

### diazona

It doesn't. Kinematics is a method of analysis that deals purely with position as a function of time, and the derivatives of that function (as opposed to dynamics, which brings in things like force and energy as well).

Nothing about kinematics says the acceleration has to be constant. It just happens that, of the realistic problems in physics that can be analyzed using pure kinematics and nothing else, most of them do involve constant acceleration, so most people only bother to write kinematic equations for constant acceleration.

14. May 22, 2010

### xxChrisxx

Oops! this is a good point acutally. It's a case of the only time I use them is in numerical simulations. So you define each step as constant acceleration so I just immediately associate the word kinematics with constant acceleration when I really shouldn't. Silly me.

So is the equation of motion a Taylor series then, or does it just look like one? As i'm not so sure any more.

15. May 22, 2010

How could it not be a taylor series? If it looks like a duck....

Seriously, do a taylor expansion of an arbitrary position function and look at the first three terms.

16. May 22, 2010

### Acut

Let me try to explain it again.
In the first post, Jack21222 showed us the equation for the motion of a constant acceleration object. This particular equation is NOT a Taylor series approximation, since it's the exact equation to describe the (idealised) motion of an object moving with constant acceleration. There are no other terms, (the d3x/dt3 term and the following ones are 0 per definition), so it's not an approximation.

HOWEVER, when you have an object subjected to all sorts of forces (i.e. the general case), than it's motion may be approximated by a Taylor series expansion - please notice the word approximated. The spring-mass system may have its motion approximated by an Taylor series (an approximation as good as your patience to calculate the following terms) but it has a simple exact solution in terms of sines and cosines.

Is it clear now?

17. May 22, 2010

### Feldoh

Yes it is, the approximation and the real solution just happen to yield the same thing. If something like this wasn't true the idea of multipole expansion could not work.

Even if you have a bunch of zeros it's still a taylor series, all the same mathematics applies to a finite exact taylor series as to the infinite taylor series.

The only requirement is something that is analytic which the kinematics equation listed by the OP is.

18. May 22, 2010