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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

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Python

Posts: 352
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
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John Gabriel wrote:
> On Wednesday, 8 November 2017 18:17:37 UTC-5, Python wrote:
>> John Gabriel wrote:
>>> On Wednesday, 8 November 2017 17:57:03 UTC-5, John Gabriel wrote:
>>>> On Wednesday, 8 November 2017 17:22:37 UTC-5, Python wrote:
>>>>> John Gabriel wrote :
>>>>>> On Wednesday, 8 November 2017 15:24:01 UTC-5, 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=0B-mOEooW03iLWldTU1ZkTDVQR0E

>>
>> 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.






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