Smallest x Value for e^(x^2): Is it Undefined?

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Discussion Overview

The discussion revolves around the mathematical exploration of the function e^(x^2) and the inquiry into the smallest x value for which this function can be defined or evaluated. Participants engage with concepts of integration, power series, and derivatives, while questioning the conditions under which the function equals zero.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents an integration of e^(x^2) and attempts to find a smallest x value by equating it to zero, leading to confusion about whether this value is undefined.
  • Another participant suggests that the integral is zero at x=0, but questions the validity of equating the series expansion to zero.
  • Some participants clarify that the function e^(x^2) is defined for all x and that there is no "smallest" x value in that context.
  • There is a discussion about the derivative of the function, with one participant stating that the minimum occurs at x=0, where the function value is 1.
  • Confusion arises regarding the purpose of substituting integer values into the series expansion and the implications of dropping the sum to form a numeric sequence.

Areas of Agreement / Disagreement

Participants express differing views on the concept of a "smallest x value" for the function e^(x^2). While some assert that the function is defined for all x, others question the reasoning behind equating the series to zero and the interpretation of the integral's behavior.

Contextual Notes

There are unresolved assumptions regarding the interpretation of the series expansion and the conditions under which the function is evaluated. The discussion also highlights the need for clarity on the relationship between the function, its integral, and its derivative.

John Sovereign
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So... I took this equation and integrated it:
e^(x^2)=1+x^2+(x^4)/2!...
∫e^(x^2)dx=(x+x^3/3+(x^5)/(5.2!)+...)+c
And then I started assigning values to x to get some weird sums. And turned them into sequences. Like for x=2
I got this sum:
2+8/3+16/5+...=S
And turned it into this:
2,8/3,16/5,...
Then I tried calculating the smallest x value for this equation, and that's where I ran into problems because you get the samething as before but this time you have to equal it to 0:

1+x^2+(x^4)/2!...=0

e^(x^2)=0

Does that mean that the smallest x value for this equation is undefined? If it isn't then what is it?
 
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May have made some mistakes! Please inform me if I have!
 
Also I didn't use erfi(x) because I don't know how to use it yet and it confuses me so instead I wrote e^(x^2) as an infinite sum.
 
John Sovereign said:
So... I took this equation and integrated it:
e^(x^2)=1+x^2+(x^4)/2!...
∫e^(x^2)dx=(x+x^3/3+(x^5)/(5.2!)+...)+c
And then I started assigning values to x to get some weird sums. And turned them into sequences. Like for x=2
I got this sum:
2+8/3+16/5+...=S
And turned it into this:
2,8/3,16/5,...
Then I tried calculating the smallest x value for this equation, and that's where I ran into problems because you get the samething as before but this time you have to equal it to 0:
You mean that the integral is 0 when x=0. That is right
1+x^2+(x^4)/2!...=0
Why? This is not the integral.
 
no i took the derrivative to calculate the minumum x value
if you take the derrivative of the integral you get the function itself.
 
That doesn't explain why you think that 1+x^2+(x^4)/2!...=0. Clearly it is =1 when x=0. The point x=0 is not the minimum of the integral. If you let x go negative, you get negative values for the integral because the integral from 0 to x has negative dx
 
Last edited:
I am completely confused as to what you are trying to do! I see that you have expanded e^{x^2} in a power series. You apparently then substituted some small integer values for x in the series, then, for some reason, dropped the sum to get a numeric sequence. Why? What did you expect to get from that?

But then, you say "Then I tried calculating the smallest x value for this equation". ? This function is defined for all x- there is NO "smallest" x value.
"that's where I ran into problems because you get the same thing as before but this time you have to equal it to 0". Do you mean you took the derivative of this function and set it equal to 0? That won't find "the smallest x value", that finds the value of x that gives the smallest function value. Is that what you meant?

Given that f(x)= e^{x^2} then the derivative is f'(x)= 2x e^{x^2} using the chain rule. No need for a power series. Since an exponential is never 0, that derivative will be 0 at x= 0 where f(0)= e^0= 1. The smallest value of f(x) occurs at x= 0 and is 1.
 

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