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Differentiation

Posted: Thu Aug 18, 2011 6:35 am
by Abdul Muntakim Rafi
I am sharing it because it seemed interesting to me...(And I haven't written any equation about calculus using the Equation editor ;) )

Don't spoil the fun for others if u already know the answer... :evil:

\[\frac{\mathrm{d} x^2}{\mathrm{d} x}= 2x\]
Okay no confusion about that...
\[\frac{\mathrm{d} x^2}{\mathrm{d} x}= \frac{\mathrm{d}(x+x+x+x+.........x times) }{\mathrm{d} x}\]
\[\frac{\mathrm{d} x^2}{\mathrm{d} x}= \frac{\mathrm{d} x}{\mathrm{d} x}+\frac{\mathrm{d} x}{\mathrm{d} x}+..................................+\frac{\mathrm{d} x}{\mathrm{d} x}\]
\[\frac{\mathrm{d} x^2}{\mathrm{d} x}= 1+1+1+1+......................+1(x ones)\]
\[\frac{\mathrm{d} x^2}{\mathrm{d} x}= x\]
Where is the mistake?

Re: Differentiation

Posted: Thu Aug 18, 2011 6:38 am
by Abdul Muntakim Rafi
it is for the beginners in Calculus(Like me)...

Re: Differentiation

Posted: Mon Aug 29, 2011 10:19 am
by prodip
No reply is coming.Will you please show the mistake?

Re: Differentiation

Posted: Tue Aug 30, 2011 8:57 pm
by Avik Roy
The first question that tickles my mind: Can you actually define $x^2$ as a sum of $x$ number of $x$'s???

However, if yo keep insisting on relying on that definition, take it this way-
The derivative of a function $f(x)$ approaches $\frac {f(x+h) - f(x)} {h}$ as $h \rightarrow 0$.
When $f(x) = x^2$, then $f(x+h) = (x+h)^2$

This can be expressed as sum of $x$ number of $x+h$'s and a residual part of $h(x+h)$. You can consider this to be a nibble of $x+h$, only $h$ out of $1$ part is taken.

And then goes $x^2$, sum of $x$ number of $x$'s. Then the subtraction $f(x+h) - f(x)$ reduces to the sum of $x$ number of $(x+h) - x = h$'s and an additional part: $h(x+h)$

Division by $h$ leaves us with the limiting value of $x$ number of $1$'s and one additional $x$. So the derivative becomes $2x$

Disclaimer: I really don't think this explanation is technically correct. It just gives an intuitive idea why the addition rule for differentiation fails when the number of terms is also variable

Re: Differentiation

Posted: Tue Aug 30, 2011 9:19 pm
by tanvirab
The addition rule fails because of countability. When we say "$x$ times", $x$ needs to be in a countable set. Then you can have a bijection with a sequence from that countable set and you stop when you reach $x$ in the sequence and call that being $x$ times. But the kind of calculus we are doing involves uncountable sets, because when you say differentiate $x^2$, you can not think $x$ to be one point, you actually need to be able to pick $x$ from an open interval around that point and that open interval is uncountable. So you can not say "$x$ times". You can only define the phrase "times" with respect to elements of a countable set.

P.S. of course you can try to define new things, that's how mathematics develops. There are theories of calculus where you don't need the notion of limit, and some of them are very algebraic. I have not read any of those yet. There are many examples of calculus on discrete and countable sets as well, and many of them are very useful in economics, physics etc. But you will need to define them in a intelligent way first.

Re: Differentiation

Posted: Sun Sep 04, 2011 1:24 am
by sm.joty
আমার মনে হয় এখানে সমস্যা টা হয়েছে এই জন্য যে এখানে আমরা ধ্রুবক আর চলকের মধ্যে ঝামেলা পাকিয়ে ফেলেছি। সমস্যাটা যেভাবে solve করেছি সেটার সাধারন সুত্র মনে হয় এই রকমঃ \[\frac{\mathrm{d} (nx)}{\mathrm{d} x}\]
\[\Rightarrow n\frac{\mathrm{d} x}{\mathrm{d} x}\]

দেখুন এখানে n=x হলে \[x^2\] এর derivative \[x\]
হয়। কিন্তু ভুলটা হল n অবশ্যই ধ্রুবক হতে হবে। কিন্তু আমরা যখন n=x লিখেছি। তখন এটা আর ধ্রুবক থাকে নি। তাই সমস্যা হয়েছে।
আমরা আসলে যেটা করেছি সেটা হল\[\frac{\mathrm{d} (x+x+x+....+x)(x times)}{\mathrm{d} x}\]
এর মানে হল\[x.\frac{\mathrm{d} (x+x+x+....+x)}{\mathrm{d} x}\]
যেটা আমার মনে হয় ঠিক না । কারন x পরিবর্তনের সাথে সাথে (x+x+x......+x) এবং (x times) দুটোই পরিবর্তিত হচ্ছে।
এই জন্যই ঝামেলা হয়েছে। :geek:

Re: Differentiation

Posted: Sun Sep 04, 2011 4:06 am
by tanvirab
Good point :)
I think my argument with countability was wrong. Need to think about it more.

Re: Differentiation

Posted: Sun Sep 04, 2011 11:47 am
by Abdul Muntakim Rafi
Avik bhaiya wrote,
The first question that tickles my mind: Can you actually define x2 as a sum of x number of x 's???
Please, explain why we can't define $x^2$ like this.

s.m. joty, Yeah... I think you got the point... Bravo! Here we are taking $x$ as a variable and constant... That's created the confusion...
And I think u made a typing mistake in the last line...

Re: Differentiation

Posted: Sun Sep 04, 2011 12:03 pm
by tanvirab
Abdul Muntakim Rafi wrote:Avik bhaiya wrote,
The first question that tickles my mind: Can you actually define x2 as a sum of x number of x 's???
Please, explain why we can't define $x^2$ like this.
Answer to that is my post above about countability :)
When we say "$x$ times", $x$ needs to be from a countable set.

Re: Differentiation

Posted: Sun Sep 04, 2011 11:24 pm
by sm.joty
Answer to that is my post above about countability :)
When we say "x times", x needs to be from a countable set.
I think Tanvir vai is right.
s.m. joty, Yeah... I think you got the point... Bravo! Here we are taking x as a variable and constant... That's created the confusion...
And I think u made a typing mistake in the last line...
'Rafi' can you say where I made a mistake. Post a quote please .