- #1
John Sovereign
- 7
- 0
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?
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?