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

  • Thread starter John Sovereign
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In summary, the conversation revolved around the integration of e^(x^2) and the use of power series and sequences to represent the function. The individual then tried to find the smallest x value for the equation, but ran into problems due to the function being defined for all x values and the derivative not giving the smallest x value but rather the x value that yields the smallest function value. The conversation also touched on the confusion around using erfi(x) and the use of the chain rule to find the derivative of e^(x^2).
  • #1
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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  • #2
May have made some mistakes! Please inform me if I have!
 
  • #3
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.
 
  • #4
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.
 
  • #5
no i took the derrivative to calculate the minumum x value
if you take the derrivative of the integral you get the function itself.
 
  • #6
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
 
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  • #7
I am completely confused as to what you are trying to do! I see that you have expanded [itex]e^{x^2}[/itex] 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 [itex]f(x)= e^{x^2}[/itex] then the derivative is [itex]f'(x)= 2x e^{x^2}[/itex] 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 [itex]f(0)= e^0= 1[/itex]. The smallest value of f(x) occurs at x= 0 and is 1.
 

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

What is the smallest possible value for x in the equation e^(x^2)?

The smallest possible value for x in the equation e^(x^2) is 0.

Why is the smallest x value for e^(x^2) considered undefined?

The smallest x value for e^(x^2) is considered undefined because the function e^(x^2) is continuous and has no minimum value. As x approaches negative infinity, e^(x^2) approaches 0 but never reaches it.

Can the smallest x value for e^(x^2) be negative?

No, the smallest x value for e^(x^2) cannot be negative because the function e^(x^2) is only defined for non-negative values of x.

Is there a limit to how small the x value can be in e^(x^2)?

No, there is no limit to how small the x value can be in e^(x^2). As x approaches negative infinity, e^(x^2) approaches 0 but never reaches it.

How does the concept of a smallest x value for e^(x^2) relate to calculus?

The concept of a smallest x value for e^(x^2) is related to calculus because it involves understanding the behavior of a continuous function, specifically in terms of its limit as x approaches negative infinity. Calculus is the branch of mathematics that deals with the study of continuous change, making it applicable to understanding the behavior of functions like e^(x^2).

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