Professor Henri Feiner and the quadratic formula

[Jeremy Weissmann]

   In 1998, I took a multivariable calculus class at my local community college as a high school senior.  The professor was Henri Feiner.  He had a thick, eastern European accent, and  —if I remember correctly—  a very dry sense of humor.

   One evening, he was working on a problem on the board, and needed to use the quadratic formula.  He applied it and started to go on with the problem.  But then he stopped and asked, “No one has any questions about what I just did?”.  It had seemed a bit funny:  It was definitely the quadratic formula or something like it.  Most of the numbers were in the right place, but not all.

   He then showed us what he had done.  He'd used a simplification that applies when the middle coefficient is even.  Indeed, if we have:

     ax² + 2bx + c  =  0     ,

then

     x  =  [ -2b ± √(4b² – 4ac) ] / 2a     .

But now a factor of  2  cancels from numerator and denominator to yield:

     x  =  [ -b ± √(b² – ac) ] / a     .

Essentially this is the quadratic formula where you take half the middle coefficient, and then use the quadratic formula "without the numbers":  just  ac  instead of  4ac ,  and just  a  instead of  2a .

   It is a very helpful simplification.  For example, if  x² + 14x - 17 = 0 ,  then we can immediately write down  x = -7 ± √(49 + 17) ,  instead of  x = [-14 ± √(14² + 68)] / 2 .   Even if we know  14² = 196 ,  we still have to work out the radicand  164 ,  only to discover that a factor of  4  is lurking, factor it out and then simplify.  But why bother if we know such a simplification can always be performed?

   And it’s not like this is some esoteric case.  As Professor Feiner pointed out, the middle coefficient will be even half the time, so this simplification will help with half the quadratics you have to solve…

   Or does it?  I have repeated that spiel for nearly 25 years, but tonight I started thinking about it more carefully.  The middle coefficient can indeed be expected to be even half the time, but the simplification will not help you with half the quadratics.

   When I realized this, I looked up Professor Feiner's phone number and called him at home.  He is 86 years old, just retired.  He was intrigued to hear the answer and we launched immediately into a discussion.  (He asked, "Were you the kid who corrected me on a series problem once?".  I said that sounded like it could have been me.)

   So the question is:  What percentage of the time will this simplification actually be helpful?

+j

[alexk]

.

    I am sorry for the late reaction, but can you give me a hint regarding the question?
   
    My problem, that i can rationalize both 0% and 100% as the valid answer for some specific cases.
   
    For example, if my goal is to solve quadratic equations numerically prof. Feiner formula is better: most coefficient wont be integer in the first place and there is no difference in cost between multiplication by 4.0 or 0.5. (I couldn't stop myself from testing it by mean of writing simple python script: the difference in their performance was about 15%)
   
    On the other hand if we are solving problems from a textbook, the author of the textbook could choose all coefficients 'b' to be odd, so we can't exploit the hack at all.
   

четверг, 1 сентября 2022 г. в 06:20:51 UTC+3, Jeremy Weissmann:

[Jeremy Weissmann]

Are you able to make sense of Feiner’s claim that it helps “half the time”?

On Jan 5, 2023, at 17:47, alexk <krae...@gmail.com> wrote:

.

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[alexk]

Hi.

As i understand, if our coefficients are integer, then half of them will be even (and half odd). My problem is that this setting is somewhat artificial, they do not have to (and in any real world application they will not) be. So are we assuming all coefficients to be random integers? Or do i miss something?

alex

суббота, 14 января 2023 г. в 18:06:20 UTC+3, Jeremy Weissmann:

[alexk]

One idea i had was to compute probability of b^2 - 4ac to be <= 0 and consider this cases as "unhelpful", but case b^2 - 4ac = 0 is statistically insignificant, b^2 - 4ac < 0  is only "unhelpful" if we exclude complex numbers. But do we?

суббота, 14 января 2023 г. в 19:10:03 UTC+3, alexk:

[Jeremy Weissmann]

Feiner’s method is definitely useful with complex solutions!

Look, the question is a bit vague. The point is, with integer coefficients, what is the fraction of the time you would actually use this method? For example, would you use it given the quadratic 2x² + 2x + 2?

On Jan 14, 2023, at 11:16, alexk <krae...@gmail.com> wrote:

One idea i had was to compute probability of b^2 - 4ac to be <= 0 and consider this cases as "unhelpful", but case b^2 - 4ac = 0 is statistically insignificant, b^2 - 4ac < 0  is only "unhelpful" if we exclude complex numbers. But do we?

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[alexk]

Well, may be? We obviously can divide all coefficient by 2, but also we can go on with it as it is.

Than again, my memory is not that great and i prefer to remember the formula, that works 100% of time.

суббота, 14 января 2023 г. в 19:18:45 UTC+3, Jeremy Weissmann:

[Jeremy Weissmann]

Interesting, I hadn’t thought about this.  Feiner’s formula would actually work just as well as dividing all coefficients by 2 and applying the standard formula.

But this wouldn’t be the case with  4x² + 4x + 4 .   So I’ll need to think about this a bit.

+j

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[MS]

Hi Team

I would like to leave this group.

My involvement was not required or suited to this forum. The common man (Me) wants insight to higher thought and seeks to know how it will solve his daily problems. However the same concepts cease to be enticing when you can't find its place in reality, or the flow is not obvious. Take care. If I can be of help please let me know.

shastry...@gmail.com

Mahesh

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[Jeremy Weissmann]

Nobody is keeping you here! 😂

On Jan 14, 2023, at 17:05, MS <msha...@acm.org> wrote:



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[MS]

Tell me how to leave, I tried, the link doesn't work

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