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## Main Question or Discussion Point

Ok, so I'm not really too good at group theory and that kind of math, so I hope I can explain my question:

I tried to evaluate [itex]\frac{d}{dx}e^{x}[/itex]:

[itex] \frac{d}{dx}e^{x} = \frac{e^{x+h}-e^{x}}{h}[/itex], h -> 0

[itex] = \frac{e^{x}e^{h}-e^{x}}{h}[/itex], h-> 0

[itex] = e^{x}(\frac{e^{h}-1}{h})[/itex], h-> 0

So I figured if I could show that: [itex] \frac{e^{h}-1}{h}[/itex] goes to 1 as h goes to 0, then the proof would be done. So I reasoned that [itex]e^{h}[/itex] goes to 1 as h goes to 0, so [itex]e^{h}-1 [/itex] must go to 0 as h goes to 0. But does [itex]e^{h}-1[/itex] "go to h" as h goes to 0?

It took me a little bit to phrase this mathematically, but I reasoned that if the difference between [itex]e^{h}-1[/itex] and h went to 0, as h went to 0, then their ratio would get closer and closer to 1, as h went to 0, and the proof would be done. So I tried it:

[itex]e^{h} - 1 - h = 1 - 1 - 0 = 0 [/itex], as h goes to 0.

So I figure that this means that [itex]\frac{e^h-1}{h} = 1[/itex], as h goes to 0. I don't know whether or not this is true, but I assume that it is because I don't see how else the derivative would end up as [itex]e^{x}[/itex]. So this brings me to my question:

So the difficult part of my problem began with two functions, which I will assume to be elements of some ring or defined on some ring (I'm not sure what ring theorists would say here), they are: [itex]e^{h}-1[/itex], and h.

I encountered a problem where the multiplicative inverse of h was multiplied to [itex]e^{h}-1[/itex]. I did not know how to take the limit of this as h went to 0.

I translated this into a problem where the additive inverse of h was added to [itex]e^{h}-1[/itex]. I DID know how to take the limit of this.

Taking the limit as h goes to 0 of this additive version of the problem, I discovered it was equal to 0. This (as far as I know) just happens to be the additive identity.

Using this information, I concluded that the limit as h goes to 0 of the multiplicative version must be 1. This (as far as I know) just happens to be the multiplicative identity.

... This is where I'm not sure how to phrase my question ... This was a nifty little trick to solve this one problem, but there seems to be a lot of structure and patterns here, and, not knowing very much group theory, I don't know what to make of it.

1) Is there some deeper thing going on here that group theory explains?

2) I.e. is this some kind of relationship between the addition part and multiplication part of a single RING?

3) Or is this like a "homomorphism" between a GROUP where the operation is addition, and another GROUP where the operation is multiplication?

4) Does this apply more generally to other groups/rings/fields, etc?

5) Is there some more general procedure to translate difficult problems involving multiplication into easier ones involving addition? How would I know when something like this can be applied?

6) Does the fact that I'm taking limits have anything to do with it?

7) Does the fact that the limit points are the additive and multiplicative identities have anything to do with it?

8) Does the continuity of the real numbers and the existence of limits have anything to do with it?

Thanks in advance for any help in understanding this

I tried to evaluate [itex]\frac{d}{dx}e^{x}[/itex]:

[itex] \frac{d}{dx}e^{x} = \frac{e^{x+h}-e^{x}}{h}[/itex], h -> 0

[itex] = \frac{e^{x}e^{h}-e^{x}}{h}[/itex], h-> 0

[itex] = e^{x}(\frac{e^{h}-1}{h})[/itex], h-> 0

So I figured if I could show that: [itex] \frac{e^{h}-1}{h}[/itex] goes to 1 as h goes to 0, then the proof would be done. So I reasoned that [itex]e^{h}[/itex] goes to 1 as h goes to 0, so [itex]e^{h}-1 [/itex] must go to 0 as h goes to 0. But does [itex]e^{h}-1[/itex] "go to h" as h goes to 0?

It took me a little bit to phrase this mathematically, but I reasoned that if the difference between [itex]e^{h}-1[/itex] and h went to 0, as h went to 0, then their ratio would get closer and closer to 1, as h went to 0, and the proof would be done. So I tried it:

[itex]e^{h} - 1 - h = 1 - 1 - 0 = 0 [/itex], as h goes to 0.

So I figure that this means that [itex]\frac{e^h-1}{h} = 1[/itex], as h goes to 0. I don't know whether or not this is true, but I assume that it is because I don't see how else the derivative would end up as [itex]e^{x}[/itex]. So this brings me to my question:

So the difficult part of my problem began with two functions, which I will assume to be elements of some ring or defined on some ring (I'm not sure what ring theorists would say here), they are: [itex]e^{h}-1[/itex], and h.

I encountered a problem where the multiplicative inverse of h was multiplied to [itex]e^{h}-1[/itex]. I did not know how to take the limit of this as h went to 0.

I translated this into a problem where the additive inverse of h was added to [itex]e^{h}-1[/itex]. I DID know how to take the limit of this.

Taking the limit as h goes to 0 of this additive version of the problem, I discovered it was equal to 0. This (as far as I know) just happens to be the additive identity.

Using this information, I concluded that the limit as h goes to 0 of the multiplicative version must be 1. This (as far as I know) just happens to be the multiplicative identity.

... This is where I'm not sure how to phrase my question ... This was a nifty little trick to solve this one problem, but there seems to be a lot of structure and patterns here, and, not knowing very much group theory, I don't know what to make of it.

1) Is there some deeper thing going on here that group theory explains?

2) I.e. is this some kind of relationship between the addition part and multiplication part of a single RING?

3) Or is this like a "homomorphism" between a GROUP where the operation is addition, and another GROUP where the operation is multiplication?

4) Does this apply more generally to other groups/rings/fields, etc?

5) Is there some more general procedure to translate difficult problems involving multiplication into easier ones involving addition? How would I know when something like this can be applied?

6) Does the fact that I'm taking limits have anything to do with it?

7) Does the fact that the limit points are the additive and multiplicative identities have anything to do with it?

8) Does the continuity of the real numbers and the existence of limits have anything to do with it?

Thanks in advance for any help in understanding this