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Topic:
Sci.math morons still struggling to prove that m+n is ALWAYS a factor of New Calculus derivative!
Replies:
1
Last Post:
Nov 8, 2017 6:32 PM



Python
Posts:
378
Registered:
8/16/16


Re: Sci.math morons still struggling to prove that m+n is ALWAYS a factor of New Calculus derivative!
Posted:
Nov 8, 2017 6:32 PM


John Gabriel wrote: > On Wednesday, 8 November 2017 18:17:37 UTC5, Python wrote: >> John Gabriel wrote: >>> On Wednesday, 8 November 2017 17:57:03 UTC5, John Gabriel wrote: >>>> On Wednesday, 8 November 2017 17:22:37 UTC5, Python wrote: >>>>> John Gabriel wrote : >>>>>> On Wednesday, 8 November 2017 15:24:01 UTC5, Python wrote: >>>>>> >>>>>> <a lot of bullshit I don't care to address> >>>>>> >>>>>>> There is actually a counter example you've always evading, >>>>>>> x > e^(1/x^2) at x = 0 >>>>>> >>>>>> That's not a counterexample you stupid! m+n is a factor of every term in the numerator. >>>>> >>>>> Mmmh, this is so obvious that you've shown it in your answer. >>>>> >>>>> Oh, dear, you didn't. >>>>> >>>>> How weird :) >>>> >>>> Oh really? Let's see how wrong you are: >>>> >>>> https://drive.google.com/open?id=1CgIaVtJjtzsp7h4SjeNfF6KdtHqzDzZA >>>> >>>> Just because YOU are not able to do it, does not mean it cannot be done. >>>> >>>> FAIL (again!) >>> >>> And just to shut your mouth: >>> >>> https://drive.google.com/open?id=0BmOEooW03iLWldTU1ZkTDVQR0E >> >> anything = (anything / (m+n) ) * (m+n), very big deal, indeed... > > Bwaaa haaaaa haaaaa!!! Love it you pathetic dog!!!! > > So you finally agree that m+n is a factor of EVERY term in the numerator eh?
sigh... Mr Gabriel, if you put m+n in the denominator of any fraction of polynomials, you'll end up with m+n in the numerator... This is silly.
Analysis deals with function on Q or R, they are not factorial rings. "being a factor of" makes no sense in this context.
So, go on, try to find the derivative at x=0. Remind: using "mainstream" calculus (with limits) not allowed.



