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May 7

Calculus

How is a double exponent integrated? Perhaps integration by substitution? What would be the solution to this? $\int {e}^{-x^{2}}\,\mathrm {d} x$ . While we're on integration by substitution, how is the value of $dx$ determined. I know in substitution it is normally replaced by $du$ , but based upon what is substituted, how is the expression for $du$ obtained? Thanks very much :), Zrs 12 (talk) 02:46, 7 May 2008 (UTC)[reply]

The integral you're talking about is the Error function - there is no simple expression for the general integral, but it does have some nice properties.

When you perform integration by substitution, you actually replace

dx

by

{\frac {dx}{du}}du

- it looks a bit like fraction cancellation, although technically it isn't. Confusing Manifestation(Say hi!) 03:29, 7 May 2008 (UTC)[reply]

What the people before me were alluding to is that

\int {e}^{-x^{2}}dx

has no elementary antiderivatives. That is using just the operations of Arithmetic (Add, Subtract, Multiply, Divide), Powers, Roots, Exponentials, Logarithms, Trigonometric Functions, and their Inverses, in any combination are incapable of describing the solution to

\int {e}^{-x^{2}}dx

. Once has to define new functions and operations to describe the solution to that integral. The Error function is once way of doing this, I suspect there are others (correct me if I am wrong). A math-wiki (talk) 04:23, 7 May 2008 (UTC)[reply]

Well, the error function is just defined from that integral, so any other thing that it evaluates to is equivalent (or a generalisation, like the hypergeometric functions). There are a few nifty properties covered at MathWorld. Confusing Manifestation(Say hi!) 04:55, 7 May 2008 (UTC)[reply]

Just a side note, if you are trying to find a definite integral, you can always use numerical methods. Or even if you want an improper integral (like from negative infinity to positive infinity) then you can also use polar coordinates.A Real Kaiser (talk) 16:48, 7 May 2008 (UTC)[reply]

Mathematica says the integral of e^(-x^2) is 0.5*pi*erf(x). If you wanted an improper integral, get the Taylor series of e^(-x^2) and integrate that. --wj32 ^t/c 08:42, 8 May 2008 (UTC)[reply]

It's

{\frac {1}{2}}{\sqrt {\pi }}\ \mathrm {erf} (x)

(note the square root). How does the Taylor series help in finding an improper integral? -- Meni Rosenfeld (talk) 09:25, 8 May 2008 (UTC)[reply]

Recursively defined sets

"A recursively defined set $C$ with a successor function of the form $S:C^{n}\to C$ , where $n$ is a positive integer, is freely generated if both $S$ is injective and the image of S does not contain any elements introduced in the basis".

Could you possibly define what "image" means in this context? For example, if the basis of $C=\lbrace 0,1\rbrace$ with $S(x,y)=x+y$ then S(0, 1) and S(1, 0) = 1, which was already defined in the basis - thus it is not freely generated- i.e., the image is all subsequently generated elements of $C$ after the basis?

Damien Karras (talk) 07:09, 7 May 2008 (UTC)[reply]

The image of a function is the subset of the range (or co-domain) which you can get to by applying the function to some element. Symbolically,

\mathrm {Im} (S)=\{a\in C\,|\,\exists b\in C^{n}\,\mathrm {st} \,S(b)=a\}=\{S(a)\,|\,a\in C^{n}\}

. In other words: yes. It's all the elements generated by the function. --Tango (talk) 12:49, 7 May 2008 (UTC)[reply]

Does this occur in one of our Wikipedia articles? It is poorly formulated. It should not be "the image of S ", but "the image of Cⁿ under S ". Also, C is a subset of some given domain D, and the function S should be typed S : Dⁿ → D. C is the limit (union) of the sequence of approximations defined by

C₀ = "the given basis";

C_k+1 = C_k ∪ S (C_k)ⁿ.

The last line uses the image notation.

The condition implies then that the relation "<" on C defined by

x < y if and only if y can be written in the form y = S (..., x, ...)

is a well-order, and furthermore that no element of the basis can be written in that way, while all other elements can, and then in a unique way. --Lambiam 14:44, 7 May 2008 (UTC)[reply]