Why Use a Dummy Variable in the Fundamental Theorem of Calculus?

[fourier jr]

What the bleep??...

i think it was in empire strikes back when obi-wan tells luke not to go try save leia/han/chewie/etc & that it's a trap, he should wait until his training is done before he faces darth vader, & yoda says "yes! to obi-wan you listen!"

edit: start at the beginning. can you tell us what is wrong (if anything) with writing
\sum_{n=0}^n f(n)

 

[Cyrus]

Sure, your index starts at n=0, and goes to n. That would be like going from 0 to 0. n only ever takes on one single value.

 

[fourier jr]

how do you fix it & explain why it works the correct way

 

[Cyrus]

I love all of you! No, honestly, I am sorry for being a pain in your rear. I am thankful for all your help! (to mathwonk)

 

[Cyrus]

Well, to make it work you would have to change the n on top of the sigma to some other value. Then n can increment from zero to, let's say r. ( if we change that n on top of sigma to r.)

\sum_{n=0}^r f(n) But I am not sure about the n inside the f(n), would it always remain at zero, or would it change? I think it would change. It would increment until it reaches the value of r and stops.

 

[matt grime]

the d3 thing. the thing is that what the person who first poted it was trying to get across was something silly. ok? no, if g is any function then the symbol dg(x) is the same as g'(x)dx. right? but the original use of this was not to refer to 3 as a constant function but to simply a number. it is moot what the intention was. I *can* give it meaning, but that isnt' necessarily what was intended. maths isn't abuot some set of things that exist and come with notation already.

 

[fourier jr]

^^ stop confusing cyrus

Well, to make it work you would have to change the n on top of the sigma to some other value. Then n can increment from zero to, let's say r. ( if we change that n on top of sigma to r.)

\sum_{n=0}^r f(n) But I am not sure about the n inside the f(n), would it always remain at zero, or would it change? I think it would change. It would increment until it reaches the value of r and stops.

yeah, & it's similar with \int_{t=a}^{x} g(t)dt

just like you can give \sum_{n=0}^r f(n) a name like F(r), we can give \int_{t=0}^{x} g(t)dt a name like G(x), where G'(x) = g(x). the 'dummy variable' t 'increments' from 0 until it reaches the 'value' x.