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Ted123
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Can anyone suggest an upper bound for [itex]e^{-x^2}[/itex] that can be integrated easily?
Ted123 said:Can anyone suggest an upper bound for [itex]e^{-x^2}[/itex] that can be integrated easily?
LCKurtz said:On what interval? Just guessing maybe ##[0,\infty)##? This is really easy. What have you thought of so far?
Ted123 said:I've just realized it's less than 1 for all x.
LCKurtz said:You haven't answered the question about what interval, or told us what problem you are trying to solve.
Ted123 said:It doesn't matter now; I know how to do it!
LCKurtz said:Not if you were trying a comparison test on ##[0,\infty)##.
No, it isn't.Ted123 said:[itex]\int_{-n}^n 1 \; dx =0[/itex]
D H said:No, it isn't.
Whovian said:Yep, it diverges, assuming you mean as [itex]\displaystyle\lim_{n\to\infty}[/itex]. Why do you think it converges?
Of course not. Given functions [itex]f(x)[/itex] and [itex]g(x)[/itex] such that [itex]0\le f(x) \le g(x)[/itex] for all x and such that [itex]\lim_{s\to\infty}\int_{-s}^s g(x) dx[/itex] diverges says nothing about the convergence or divergence of [itex]\lim_{s\to\infty}\int_{-s}^s f(x) dx[/itex].Ted123 said:Oh yeah, it equals [itex]2n[/itex] which diverges as [itex]n\to\infty[/itex].
But doesn't that contradict the comparison test?
D H said:Of course not. Given functions [itex]f(x)[/itex] and [itex]g(x)[/itex] such that [itex]0\le f(x) \le g(x)[/itex] for all x and such that [itex]\lim_{s\to\infty}\int_{-s}^s g(x) dx[/itex] diverges says nothing about the convergence or divergence of [itex]\lim_{s\to\infty}\int_{-s}^s f(x) dx[/itex].
There are plenty of positive definite functions [itex]f(x)[/itex] that are bounded from above by [itex]g(x)=1[/itex] whose integral over the real number line does converge. [itex]f(x)=\exp(-x^2)[/itex] is one such function.
No. That would be solving your homework for you.Ted123 said:OK, so I need an upper bound whose integral converges then. Any suggestions?
Mentallic said:What about something with [tex]xe^{-x^2}[/tex]
D H said:Do you care? exp(-x^2) is symmetric, so if the integral from 0 to infinity exists, then so does the integral from -infinity to infinity.
Ted123 said:But if [itex]x<0[/itex] then [itex]e^{-x^2} > xe^{-x^2}[/itex]
An upper bound for e^(-x^2) is the largest possible value that the function can reach. It is often used in mathematical analysis to determine the behavior of a function and its limit.
The upper bound for e^(-x^2) is typically calculated using mathematical proofs and techniques, such as the Mean Value Theorem or the Cauchy-Schwarz inequality. It is important to note that the upper bound may vary depending on the interval or range of values for x.
The Gaussian function, also known as the bell curve, is closely related to the upper bound for e^(-x^2). The Gaussian function can be represented as e^(-x^2/2), which is a special case of e^(-x^2). The upper bound for e^(-x^2) is often used to analyze the behavior of the Gaussian function.
No, an upper bound for e^(-x^2) cannot be negative. The function e^(-x^2) is always positive, and its upper bound represents the maximum value it can reach. If the upper bound is negative, it would contradict the nature of the function.
An upper bound for e^(-x^2) is useful in various fields, such as statistics, physics, and economics. It can help in approximating the behavior of a system or predicting the outcomes of experiments. In economics, it is used to analyze the distribution of income or wealth among a population.