1. Aug 24, 2015

### John Sovereign

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?

2. Aug 24, 2015

### John Sovereign

3. Aug 24, 2015

### John Sovereign

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. Aug 24, 2015

### FactChecker

You mean that the integral is 0 when x=0. That is right
Why? This is not the integral.

5. Aug 25, 2015

### John Sovereign

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. Aug 25, 2015

### FactChecker

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: Aug 25, 2015
7. Aug 25, 2015

### HallsofIvy

Staff Emeritus
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.