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Topic: John Gabriel's Thread on Mathematics.
Replies: 231   Last Post: Mar 22, 2014 9:23 PM

 Messages: [ Previous | Next ]
 David C. Ullrich Posts: 3,555 Registered: 12/13/04
Re: New Calculus for Dummies.
Posted: Mar 1, 2014 12:20 PM

On Fri, 28 Feb 2014 10:48:52 -0800 (PST), iwannig@gmail.com wrote:

>On Friday, February 28, 2014 5:04:08 PM UTC+2, dull...@sprynet.com wrote:
>

>> Whee! You need to use the power series for this? Wow.
>
>Nope. I didn't have to. I could have used the Gabriel Polynomial which is finite, but most of you have never seen it. There is no problem using a power series.

I asked how you find the derivative of sin(x). You showed me a proof
using the power series. Now when I ask the obvious questions about
that you say you didn't have to. You should either answer my
questions or _show_ me the proof without power series.

>
>> See, I thought the point was the New Calculus was going to avoid
>> limits. How in the world do you make sense of a power series
>> without limits.

>
>Very easily. In most cases, you NEVER know the limit of the sine series. It's always an approximation. The limit is not required at all in calculus. In fact, It has no place in calculus.
>

>> Heh, a better question. You seem to think that 1 is not equal to
>> 0.999...

>
>I *know* it's not equal to 1. There's no doubt about that. Not a single proof exists to support the fallacy besides Euler's blunder (defining the limit to be the infinite sum, even when there is no such thing as an infinite sum).
>

>> Now, saying 1 = 0.999... is the same as saying that
>> 1 is equal to a certain infinite series:
>> (*) 1 = .9 + .09 + ... .

>
>That's not true. It was defined that way. That's the only support for the assertion.
>

>> You have problems with (*), but the much more sophisticated
>> series
>> (**) sin(x) = x - x^3/3! + x^5/5! ...
>> is no problem?

>
>I do not accept infinite series. The above series is NOT infinite.

Not infinite except for being an infinite series, fine.

>There is no such thing as an infinite series. Everything I need to know about the series is contained in the first three terms. :-)

Huh? Then show me the proof that just uses the first three terms.
Also the proof that you have the fiirst three terms right. As well
as an expllanation of what you mean by that...

(Because all I can think of as an explanation for why the first
three terms are right is that the sum of those terms agrees
with sin(x) up to a certain number of derivatives at the
origin. But you can't say _that_ since you don't even know
what the derivative of sin(x) _is_ yet...)

>
>

>> Seems to me that whatever reason you have for saying that 0.999... does not equal 1 exactly should also say that x - x^3/x! + ... does not equal sin(x) exactly.
>
>I'd say that's an accurate assessment! :-)
>

>> What's the diifference?
>
>None.
>
>Any power series is always an approximation. Unless of course there are other methods of determining the series given a particular value.
>
>The New Calculus does not use limits, infinity or infinitesimals in any shape or form.
>
>The power series that you saw used in the derivation happens to equal to 0 because all the terms except for the first term contain the factor m+n.
>
>While you can do this with 0, you can't do it with any other value.

Date Subject Author
2/22/14 thenewcalculus@gmail.com
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3/1/14 David C. Ullrich
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3/6/14 David C. Ullrich
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3/3/14 Roland Franzius
3/3/14 thenewcalculus@gmail.com
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3/3/14 Roland Franzius
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