How 3=2 is true?,help me to find out the mistakes

[Nero26]

m^2+m^2+m^2+...up to m term =m^2*m=m^3
Which is identity .So we use derivative for this identity.
Now, d/dm(m^2+m^2+m^2+...up to m term)=d/dm(m^3)
Or 2m+2m+2m+...up to m term=3m^2
Or 2m*m=3m^2
Or 2m^2=3m^2 but it is not possible.So there should be some mistakes.But I'm not getting it.
What I suspect are as follows:
In L.H.S we differentiated first,then add all terms.In R.S we added all then differentiated it.Is the mistake for maintaining different orders of operation?
I suspect another reason like while differentiating R.S, number of terms is considered as variable and multiplied with m^2 that leads to m^3,but as it is no. of terms if I assume it as constant then, m*d/dm(m^2)=m*2m=2m^2,which matches with L.S. Is this okay?or something else?
Thanks for your help.

 

[foxofdesert]

Here is my thought:
you have m-terms of m^2. That is, m*m^2. Taking derivative, d\dm (m*m^2)=m^2+m*2m=3m^2.
The problem with d/dm(m^2+m^2+m^2+...up to m term) is that m-term is also a function of m. So, technically, you need to differentiate the function (m-term)

 

[Ray Vickson]

m^2+m^2+m^2+...up to m term =m^2*m=m^3
Which is identity .So we use derivative for this identity.
Now, d/dm(m^2+m^2+m^2+...up to m term)=d/dm(m^3)
Or 2m+2m+2m+...up to m term=3m^2
Or 2m*m=3m^2
Or 2m^2=3m^2 but it is not possible.So there should be some mistakes.But I'm not getting it.
What I suspect are as follows:
In L.H.S we differentiated first,then add all terms.In R.S we added all then differentiated it.Is the mistake for maintaining different orders of operation?
I suspect another reason like while differentiating R.S, number of terms is considered as variable and multiplied with m^2 that leads to m^3,but as it is no. of terms if I assume it as constant then, m*d/dm(m^2)=m*2m=2m^2,which matches with L.S. Is this okay?or something else?
Thanks for your help.

You are not allowed to take a derivative of a sum like $S(m) = \sum_{i=1}^m a_i$ with respect to m. Think about it: if S(m) had a derivative, it would be
$$S'(m) = \lim_{h \to 0} \frac{\sum_{i=1}^{m+h} a_i - \sum_{i=1}^m a_i}{h} = \lim_{h \to 0} \frac{\sum_{i=m}^{m+h} a_i}{h}.$$ If you take, for example, h = 0.001, what on Earth could you possibly mean by $\sum_{i=m}^{m + 0.001} a_i?$ It makes no sense.

 

[Nero26]

Thanks to all for your help.

You are not allowed to take a derivative of a sum like $S(m) = \sum_{i=1}^m a_i$ with respect to m. Think about it: if S(m) had a derivative, it would be
$$S'(m) = \lim_{h \to 0} \frac{\sum_{i=1}^{m+h} a_i - \sum_{i=1}^m a_i}{h} = \lim_{h \to 0} \frac{\sum_{i=m}^{m+h} a_i}{h}.$$ If you take, for example, h = 0.001, what on Earth could you possibly mean by $\sum_{i=m}^{m + 0.001} a_i?$ It makes no sense.

I couldn't understand this line,
$\sum_{i=1}^{m+h} a_i - \sum_{i=1}^m a_i=\sum_{i=m}^{m+h} a_i$
$\sum_{i=m}^{m+h} a_i=a_m+a_{m+1}+a_{m+2}+...+a_{m+h}$ (assuming h is some integer)But terms upto $a_m$ should be canceled by $\sum_{i=1}^{i=m}a_i$ then what remains is $a_{m+1}+a_{m+2}+...+a_{m+h}$ which is $\sum_{i=m+1}^{i=m+h}a_i$ .Now in this case as h→0, after subtraction we'll get a term $a_{m+h}$ that fails to exist.
Another thing I couldn't understand is:
$$S(m)=\sum^{i=1}_{i=m}a_i$$,Whether a is constant or function of m?
However ,it seems I got your point that as number of terms can't be a fraction so we can't get derivative,like $a_{m+0.001}$ can't exist.Am I right?

 

[LCKurtz]

Whew! Glad to see that's settled, because it is well known that 3 = 2 only for a large value of 2.

 

[someGorilla]

R: How 3=2 is true?,help me to find out the mistakes

Your left hand term is only defined for integer values of m, so d/dm makes no sense.

 

[vela]

Whew! Glad to see that's settled, because it is well known that 3 = 2 only for a large value of 2.

Yes, order has been restored. We can all sleep easy tonight!

 

[Mark44]

Whew! Glad to see that's settled, because it is well known that 3 = 2 only for a large value of 2.

Or a very small value of 3...