[see also: moment, simultaneously
an impulse acting at time t=0
This path stays in B for all time.
This can be performed in O(n) time.
The problem is to move all the disks to the third peg by moving only one at a time.
At times [= Occasionally] it will be useful to consider......
We prove both lemmas at the same time.
At the time of writing , I was not aware of this reference.
It has been known for some time that......
But this time boundedness on U is enough; we do not need continuity on V.
Then A has three times as many elements as B has.
The integral of F is r times the sum of......
Thus X has r times the length of Y.
The diameter of A is three times that of B.
Clearly, F is r times as long as G.
Applying this argument k more times, we obtain......
......where zk=z... z (k times)
Let K be the number of times that z returns to B.