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Taylor Series Expansion


Date: 11/24/2001 at 15:14:59
From: Don Castaldini
Subject: Taylor Series Expansion

The distance from A to B is 1000 meters. As one traverses the distance 
from A to B, at 1 meter per second, the distance is instantaneously 
and uniformly stretched 1000 additional meters. How long does it take 
to get from A to B (in seconds)?

What we've figured is that as we take one step toward B, the road 
stretches 999 meters in front and 1 meter behind. This happens each 
time we take a step toward B.

What we've tried to do is come up with a formula for what fraction of 
the total distance has been covered after n seconds, but just before 
the distance stretches; however, we've been unable to come up with 
anything that isn't crazy looking!


Date: 11/25/2001 at 15:23:38
From: Doctor Pete
Subject: Re: Taylor Series Expansion

Hi,

This is an interesting problem, one that depends on whether you 
consider it the distance discrete or continuous. What do I mean by 
this? If the problem is discrete (as implied by the wording, and your 
comments), then each time we step 1 meter forward, the length 
stretches 1000 meters under our feet. On the other hand, if the 
problem is continuous, then you can imagine we're on a bicycle 
instead, moving at a constant rate of 1 m/s, and the road beneath us 
is stretching at a smooth, continuous rate of 1000 m/s.

The important thing to realize is that the two situations I have thus 
described are different, and will give different answers. Let us 
consider the discrete case first. 

At time t, let x[t] be the distance from point A, and d[t] be the 
total distance from A to B.  Then clearly,

     d[t] = 1000(t+1).

The other thing to notice is that at time t+1, we will have traveled 
1 meter farther than x[t], plus an amount corresponding to the 
stretching of the road beneath us. While the amount of stretch is the 
same (1000 m), how far it moves us from x[t] depends on how far we 
have traveled. In particular, this additional stretch moves us a 
distance of

     1000(x[t]+1)/d[t],

assuming that the road stretches after we step 1 meter forward from 
x[t]. Therefore, we find the distance

     x[t+1] = x[t] + 1 + 1000(x[t]+1)/d[t]
            = (x[t] + 1) + (x[t]+1)/(t+1)
            = (x[t] + 1)(1 + 1/(t+1)),

or equivalently,

     x[t] = (x[t-1] + 1)(1 + 1/t),
     x[0] = 0.

You may verify that at time t = 1, we have traveled a distance of

     x[1] = (x[0] + 1)(1 + 1/1) = 2.

Now, the solution to this recursion relation is not easy. My best 
explanation is to say that the solution is

     x[t] = (t+1)H[t],

where

     H[t] = 1 + 1/2 + 1/3 + 1/4 + ... + 1/t

is the t(th) harmonic number. (For convenience, let H[0] = 0). To 
verify that this is the solution, we simply substitute it into the 
right-hand side of the recursion relation, giving

     (tH[t-1] + 1)(1 + 1/t) = (tH[t-1] + t/t)(1+1/t)
                            = tH[t](1+1/t)
                            = tH[t] + H[t]
                            = (t+1)H[t],
                            = x[t],
as claimed.

Finally, since x[t] gives the position from A at time t, and d[t] 
the total distance from A to B, we seek the value of t > 0 that gives 
x[t] = d[t], or

     (t+1)H[t] = 1000(t+1),
or
     H[t] = 1000.

This is not easy to compute, since t is very large. Using the relation

      lim [H[t] - ln[t]] = g,
     t->oo

where g is the Euler-Mascheroni constant 0.5772156649015328606...., we 
have approximately

     1000 - ln[t] = g,
or
     t = Exp[1000 - g],
or
     t = 1.106115110266 x 10^434,

The total distance traveled is 1000(t+1), which is essentially the 
same as 1000t, which is 1.106 x 10^437 meters.


The continuous case is actually quite a bit simpler. As before, we 
have

     d[t] = 1000(t+1),

but now x[t] satisfies the differential equation

     x'[t] = 1 + x[t]/(t+1)

(with initial condition x[0] = 0), because the speed at time t is 
1 m/s plus the rate at which the ground behind us is stretching, which 
is 1000x[t]/d[t]. Since it is linear, the solution is simply

     x[t] = (t+1)(C + ln[t+1]);

the initial condition x[0] = 0 implies C = 0, and hence

     x[t] = (t+1)ln[t+1].

Therefore, x[t] = d[t] implies ln[t+1] = 1000, or

     t = Exp[1000] - 1

which is approximately 1.970071114017 x 10^434. Note the similarity of 
the solutions.  

- Doctor Pete, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Calculus
High School Calculus
High School Sequences, Series

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