About Palmer's "Cadillac" Quadratic Solver



#5

Palmer,

I was just reading your Article 593, and working the examples for the 8 cases.

I noticed that case 7 and 8 in this article are the same as the two additional cases in your Article 396, but that the result for R2 in the first additional case in Article 396 doesn't match the result for case 7 in Article 593.

Also, the result for the imaginary part of the roots of the second additional case in Article 396 doesn't match the result for case 8 in Article 593.

The error seems to be in the Article 396 results, and in fact, the R2 result for the first additional case in Article 396 seems to be the square root of the correct result.


#6

Rodger:

You are correct that I have published different results for the same problems. The situation is even worse than you describe as it also goes back to the documentation of the extended quadratic for the HP-41 which was where I started playing around with the funny properties. At the moment I don't know why I did that but I have started digging through my notes. I should be able to sort it out by the weekend.

Once again, all I can offer is a mea culpa.


#7

I agree that the errors are in the Article 396 results.

Looking at the first additional case in Article 396:

If a = 11,111,119 ; b= 11,111,111 ; c = 11,111,103

Then d = 123,456,78,654,321 - 123,456,786,257 = 64

Since d>0

R1 = (b + sqrt(d))/a = (11,111,111 + 8)/11,111,119 = 1

R2 = (b - sqrt(d))/a = (11,111,111 - 9)/11,111,119 = 0.999998560001

The value 0.999999280001 = b/a = 11,111,111/11,111,119

which is the value for both R1 and R2 if the word length of the machine in use is such that d is calculated as zero.

The correct value of R2 from Article 593 is close to, but not equal to, the square of the incorrect value of R2 in Article 396 where the difference is d/a2

Of course, that doesn't explain how I managed to publish the incorrect value for R2 in Article 396.

I have yet to figure out where the incorrect value for the imaginary part of the second additional problem came from.


#8

Quote:

I have yet to figure out where the incorrect value for the imaginary part of the second additional problem came from.


Now I have:

If you divide 8 by 11,111,119 you will get the value shown in Article 396 for the imaginary part of the answer for the second additional test case. Unfortunately, those are the values for the square root of d and a from the first additional test case!

As a registered Republican I feel comfortable in saying that I can't recall how I did that.


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