Problem: Place four numbers in the first row as follows: A B C D. For each successive row replace the entries by the absolute value of the difference of the entry just above and the entry just to the right in the previous row. In the fourth position use the absolute vale of the difference of the fourth and the first (i.e. cycle). abs.A-B, abs B-C, abs C-D, and abs D-A.
A.) Will the process lead to a 0 in all 4 entries for some row?
B.) What is the largest number of rows before a zero row is generated?
Hint: If answer is less than 10, you should try again.
To answer part a, the process will lead to a 0 in all four entries. The only difference is where that row of zeros appear. What makes the difference is that by taking the absolute value you will never have negative numbers and there will always be a way to get a row of all zeros. For example, let's start with 1, 2, 3, and 4. Then I followed the steps as mentioned above.
As you can see, the sixth row is a row of zeros. As you notice, we will get a row of zeros when all four entries become the same number. This would be the only way to get a row of zeros.
After this I tried increasing the values by increments of one, but did not get very far. I began trial and error to see what types of conclusions I could come up with. I knew that I had to keep at it until I at least came up with four numbers that gave me more than 10 nonzero rows. Two examples that I came up with follow below.
and the following is a second example that I came up.
In these two example, you can see that I came up with four numbers that give me thirteen nonzero rows and therefore the fourteenth row is zeros.
Problem : Explore problems of maximization such as the lidless box formed from a 5x8 sheet with a square removed from each corner.
You can see from the above table that the maximum volume is when x=1 the volume is 18.
Problem: Find the maximum for f(x) = (1-x)(1+x)^2 on the interval [0,1].
From the above table so far we see that the maximum is going to be somewhere between .25 and .50.
I could have done more values in my spreadsheet that were between .25 and .50, but it is easy to see that the value (from Calculus) was 1/3.