Prometheus Posted November 11, 2016 Posted November 11, 2016 Apparently [latex]xe^{-x^2}[/latex] is an odd function. I found out the long way finding the integral sums to zero (btw, excluding trivial cases, is that a sufficient condition to conclude a function must be odd?). However, i am supposed to be able to tell this was odd by inspection (similarly for even functions). I'm sure they don't mean i need to be able to determine the parity of any function by inspection, but does anyone know of any hints and tricks for doing so? I'm starting to see a pattern where the parity of a Gaussian integral multiplied by a polynomial is equal to the parity of the polynomial. Any others?
imatfaal Posted November 11, 2016 Posted November 11, 2016 I would break it down and grok what bits do what for different portions of the number line; ie big negative, fractional negative, zero, fraction, big positive x^2 is always positive, so -x^2 is always negative so e^(-x^2) is symmetrical about y axis the multiplication by x means that the expression will change in magnitude by the same amount on each side of the y axis BUT that the sign will be flipped for negative x. ie f(-x) will equal -f(x)
studiot Posted November 11, 2016 Posted November 11, 2016 (edited) I found out the long way finding the integral sums to zero (btw, excluding trivial cases, is that a sufficient condition to conclude a function must be odd?). No. The definition of an odd function is f(x) = -f(-x) The definition of an even function is f(x) = f(-x) That makes the zero function even, but the integral of the zero function over any interval is zero. There is no such thing as a trivial case in proofs. Edited November 11, 2016 by studiot
fiveworlds Posted November 11, 2016 Posted November 11, 2016 so -x^2 is always negative Nope where x = root(y) =( -y^1/2)^2 = -y^(2/2) = -y
Strange Posted November 11, 2016 Posted November 11, 2016 Nope where x = root(y) =( -y^1/2)^2 = -y^(2/2) = -y That doesn't appear to have anything to do with the question. If x = -y then x2 = y2 therefore -x2 is negative. x2 is always negative (unless x is complex).
uncool Posted November 11, 2016 Posted November 11, 2016 I would break it down and grok what bits do what for different portions of the number line; ie big negative, fractional negative, zero, fraction, big positive x^2 is always positive, so -x^2 is always negative so e^(-x^2) is symmetrical about y axis the multiplication by x means that the expression will change in magnitude by the same amount on each side of the y axis BUT that the sign will be flipped for negative x. ie f(-x) will equal -f(x) I think you've mixed up a bit; "always positive" and "always negative" are irrelevant here. For example, e^x is always positive, but isn't even. What's relevant is: x^2 is even (symmetrical about y-axis) so -x^2 is even so e^(-x^2) is even x is odd so x e^(-x^2) is odd.
Xerxes Posted November 11, 2016 Posted November 11, 2016 I am not quite sure what's going on here - perhaps some are conflating the notion of odd/even numbers with that of odd/even functions. Here's a (sort of) definition: A function is even if, for every element in its domain, its sign id preserved in the codomain (sometimes called the "range") Example: if [math]f(x) = 2x[/math] for all Real [math]x[/math], then obviously [math]f(x)+f(-x)=0[/math] Conversely, a function is odd if,for every element in the domain it reverses the sign in the codomain. Example: if [math]f(x)=x^2[/math] for all Real [math]x[/math], then obviously [math]f(x)+(-(f(-x))=f(x)-f(-x)=0[/math]. What's hard about that?
studiot Posted November 12, 2016 Posted November 12, 2016 (edited) xerxes A function is even if, for every element in its domain, its sign id preserved in the codomain (sometimes called the "range") You might like to consider if you have this the wrong way round. Consider f(x) = x3. If x = -3 then f(x) = -27 so the sign is preserved. Is x3 odd or even? Edited November 12, 2016 by studiot
Xerxes Posted November 12, 2016 Posted November 12, 2016 Is x3 odd or even?I don't understand the question. [math]x^3[/math] is a number, and is odd or even according to whether [math]x[/math] is odd or even. the function [math]f: \mathbb{R} \to \mathbb{R}, x \mapsto x^n[/math] is even or odd according to whether [math]n[/math] is odd or even,
studiot Posted November 12, 2016 Posted November 12, 2016 (edited) I don't understand the question. [math]x^3[/math] is a number, and is odd or even according to whether [math]x[/math] is odd or even. the function [math]f: \mathbb{R} \to \mathbb{R}, x \mapsto x^n[/math] is even or odd according to whether [math]n[/math] is odd or even, The question refers to the quote taken from your previous post and highlighted in my post # 8 In your terminology, x is not a number it is an element of the domain and x3 is not another number, it is an element of the codomain. You are the one who chose to introduce this method of describing a function. The point I am making is that the sign of the element (-3 in this case) in the domain is the same as the sign of the element in the codomain (-27 in this case), which was your specified definition of an even function in your post#7. This contradicts your statement in your post#9 which states that if n = 3, and therefore odd, the function is odd, since 3 is odd. As a matter of interest the range and the codomain do not necessarily coincide for a function. Edited November 12, 2016 by studiot
Xerxes Posted November 12, 2016 Posted November 12, 2016 This contradicts your statement in your post#9 which states that if n = 3, and therefore odd, the function is odd, since 3 is odd.I said no such thing. Read again, paying particular attention to word ordering
studiot Posted November 12, 2016 Posted November 12, 2016 (edited) I said no such thing. Read again, paying particular attention to word ordering Yes indeed I missed the order reversal so your two posts are consistent. Sadly they are consistently wrong. Edited November 12, 2016 by studiot
John Cuthber Posted November 12, 2016 Posted November 12, 2016 You say "Apparently is an odd function. I found out the long way finding the integral sums to zero"That's interesting.If I wanted to find if it was odd or even I'd split it into two parts.x and e^-x2y=x is clearly an odd function. and y= e^-x2 is a well known function- give or take a scaling factor- it's the normal distribution and it's symmetrical so it's even Here's the important bit- the product of an even function and an odd function is always odd and that answers the question for you. But I'm more interested in the integration you did. There is no analytic form for the integral of the normal distribution- that's why there are lots of table for it, but you have integrated a very similar function and I'd like to know how.
studiot Posted November 12, 2016 Posted November 12, 2016 (edited) You say "Apparently is an odd function. I found out the long way finding the integral sums to zero" That's interesting. If I wanted to find if it was odd or even I'd split it into two parts. x and e^-x2 y=x is clearly an odd function. and y= e^-x2 is a well known function- give or take a scaling factor- it's the normal distribution and it's symmetrical so it's even Here's the important bit- the product of an even function and an odd function is always odd and that answers the question for you. But I'm more interested in the integration you did. There is no analytic form for the integral of the normal distribution- that's why there are lots of table for it, but you have integrated a very similar function and I'd like to know how. I haven't worked it fully through, but does not the substitution z = x2 dz = 2xdx [math]I = \frac{1}{2}\int {{e^{ - z}}} dz[/math] Not suffice? Edited November 12, 2016 by studiot
John Cuthber Posted November 13, 2016 Posted November 13, 2016 "In fact, there is no exact "closed form" expression for the value of this integral over an arbitrary range. " from Integrating The Bell Curve http://www.mathpages.com/home/kmath045/kmath045.htm
Prometheus Posted November 13, 2016 Author Posted November 13, 2016 Thanks for all the tips, i find just being aware helps a great deal. . . . That makes the zero function even, but the integral of the zero function over any interval is zero. There is no such thing as a trivial case in proofs. I thought i was using the term correctly, for instance this wikipedia page refers to the zero function as trivial. Does it depend on context? But I'm more interested in the integration you did. There is no analytic form for the integral of the normal distribution- that's why there are lots of table for it, but you have integrated a very similar function and I'd like to know how. Its not quite a normal distribution though (integrated over the entire set of reals, which is what i did, its the expectation of a normal), its multiplied by x which allows the substitution studiot gave to work.
studiot Posted November 13, 2016 Posted November 13, 2016 Thanks for all the tips, i find just being aware helps a great deal. I thought i was using the term correctly, for instance this wikipedia page refers to the zero function as trivial. Does it depend on context? Its not quite a normal distribution though (integrated over the entire set of reals, which is what i did, its the expectation of a normal), its multiplied by x which allows the substitution studiot gave to work. What is trivial does indeed depend upon the context. Remember it only takes one except to disprove the most elegant theory in maths. The zero function is y = 0 for all x. This is not quite the same as for instance y = ax3 which is odd for all a except a = 0, when it is neither odd nor even. It is a subtle point and I wanted to move on from towards noting that in general functions are neither odd nor even, but may be decomposed into the sum of an odd function and an even function. Odd and even have significance in science, in that they are examples of a wider phenomenon known as parity. Even functions have parity +1 and odd functions have parity -1 Parity is related to symmetry, so an even function is symmetric with respect to the y axis, x = 0 and an odd function is antisymmetric with respect to the x axis, y = 0
Function Posted November 13, 2016 Posted November 13, 2016 I'm a bit odd ... But would that make me an odd Function? Sorry, couldn't resist
studiot Posted November 13, 2016 Posted November 13, 2016 I'm a bit odd ... But would that make me an odd Function? A bit odd?? But what about the other bit ie the rest of you? So you are really like most functions, the sum of an odd and an even function.
Function Posted November 13, 2016 Posted November 13, 2016 (edited) A bit odd?? But what about the other bit ie the rest of you? So you are really like most functions, the sum of an odd and an even function. Okay, we're even now. There's the other bit; the even Function. But if I am like most other functions, would that make almost every Function a bit odd? Edited November 13, 2016 by Function 1
studiot Posted November 13, 2016 Posted November 13, 2016 (edited) Okay, we're even now. There's the other bit; the even Function. But if I am like most other functions, would that make almost every Function a bit odd? Touche +1 Edited November 13, 2016 by studiot
wtf Posted November 13, 2016 Posted November 13, 2016 But if I am like most other functions, would that make almost every Function a bit odd? Yes, just like 2 is a very odd prime.
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