there is no real number for that equation.
as lim x ==> oo, e^x = 0
Soriku said:
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It's not a solution: e^(something big)=(something bigger), thus e^(positive infinity)=(positive infinity).
Are you familiar with limits? That's the only way these statements make sense in standard analysis.
| Soriku said: lol, no one helped me last time.
Almighty math wizard, what does x equal in e^x = 0. I can't figure it out :x |
lol...I'm not a math wizard...those questions were pretty easy imo....
Soriku said:
Uh, I don't think we're that far yet. |
The rigth answer is that no number exists to satisfy e^x = 0.
"Minus infinity" is just the limit to which x goes when the expression (e^x) approaches zero. But you can never really get it to equal zero.
Think of it this way, the smaller the x you use, the smaller and closer to 0 e^x will be, but it'll never, ever, actually get there. So you can get as close to 0 as you want by using a "big" enough negative number even if you can never quite reach 0. That's why the notion of "limits" is used. 0 would be the limit of e^x as x approaches minus infinity.
But it's by no means a number that solves your equation. If you're at highschool or something like that, I think it would be more right to simply say such a number doesn't exist, or does not pertain the Real set, or that the solution set is empty, or anything like that.
Sorry about all the redundancy. Oh, and the questions are actually pretty easy if you know what you're doing :P
Soriku said:
Yeah, that's what I figured...I'll add a note to my HW saying that but I think negative infinity sounds accurate too. |
Well, The idea of negative infinity equaling to zero there is the same as the idea of n/0 equaling infinity, and most math teachers are pretty adamant in saying you can never, ever, for your life, divde by zero :P
| Farmageddon said:
|
A preamble:
I agree that if Soriku hasn't even reached limits in his studies, the context is probably that of real numbers - thus the correct answer is "there's no solution of the equation in the real number set".
A digression:
that said, you're not correct in the first part of your statement. If you extend real numbers to hyperreal numbers you have a rigorous and coherent way to treat infinite and infinitesimal numbers.
In R* (the hyperreal set) if dx is an infinitesimal, for example, 12/0 keeps having no meaning but 12/dx has (it's an infinte quantity, of course, and thus you can't take its standard real part).
Thus in such context division by zero keeps having no defined meaning, but you can formulate in a "static" way some expressions that require limits in standard analysis and calculus, such as Soriku's equation.
| GodOfWar_3ever said: 1) If the initial # of members = x 720/x = y ----> From this, 720 = xy 720/(x-2) = [y (plus) 4] ----> From this 720 = (x-2)[y(plus)4] = xy (plus) 4x - 2y - 8 Since they're both equal to 720, xy = xy (plus) 4x - 2y - 8 0 = 4x - 2y - 8 2y = 4x - 8 y = 2x - 4 Substitute this in 720 = xy and we get, 720 = x(2x-4) 720 = 2x2 - 4x Divide everything by 2, 360 = x2 - 2x x2 - 2x - 360 = 0 (x plus 18)(x - 20) = 0 Therefore, x = -18 or x=20 Since -18 cannot be taken, the # of members is 20 before, 18 after. 2) Don't know if you've done this at school but to check the roots of a quadratic equation, you can use this thing called the discriminant (Think thats what its called) Discriminant = (coefficient of the "x" term)2 - 4(coefficient of the x2 term)(coefficient of the constant term) If its equal to 0, then the equation has real, equal roots. If its more than 0, then there are 2 real, different roots. If its less than 0, the roots are not real... I think Anyway, we have to consider the discriminant to be equal to 0. Coefficient of the, x term - -5 x2 term - p constant term - p Therefore, for the equation to have real roots, (-5)2 - 4(p)(p) = 0 25 - 4p2 = 0 25/4 = p2 p = 5/2 or p = -5/2 3) I don't understand what you're saying.... I'm gonna assume one root is 2 plus 3 sqr.5 and the other is 2 - 3 sqr.5 So the equation is - (x-[2 plus 3sqr.5])(x-[2 - 3sqr.5]) = 0 x2 -[2 - 3sqr.5]x - [2 plus 3sqr.5]x plus [2 plus 3sqr.5][2 - 3sqr.5] = 0 x2 - 2x plus 3sqr.5x - 2x - 3sqr.5x plus (22 - 3^2(sqr.5)^2) = 0 (^2 = squared btw) x2 - 4x -4 plus 4 - 45 = 0 x2 - 4x - 41 = 0 God I hope that correct.... 4) If the boat is travelling at x km/h, [65/(x - 5)] plus [55/(x plus 5)] = 15 [65(x plus 5) plus 55(x - 5)] that whole thing divided by (x plus)(x-5) = 15 Divide both sides by 5 to get [13(x plus 5) plus 11(x - 5)] that whole thing divided by (x plus)(x-5) = 3 [13x plus 65 plus 11x - 55] divided by x2 - 25 = 3 24x plus 10 = 3x2 - 75 3x2 - 24x - 85 = 0 x2 - 8x - 85/3 = 0 x2 - 8x plus 16 - 133/3 (x - 4)2 = 133/3 x - 4 = sqr.(133/3) x - 4 = 6.658 x = 10.658 There are bound to be some mistakes cause typing this shit in and no plus symbols is pain in the ass |
Thanks so much ;)
WereKitten said:
A preamble: I agree that if Soriku hasn't even reached limits in his studies, the context is probably that of real numbers - thus the correct answer is "there's no solution of the equation in the real number set". A digression: that said, you're not correct in the first part of your statement. If you extend real numbers to hyperreal numbers you have a rigorous and coherent way to treat infinite and infinitesimal numbers. In R* (the hyperreal set) if dx is an infinitesimal, for example, 12/0 keeps having no meaning but 12/dx has (it's an infinte quantity, of course, and thus you can't take its standard real part). Thus in such context division by zero keeps having no defined meaning, but you can formulate in a "static" way some expressions that require limits in standard analysis and calculus, such as Soriku's equation.
|
If you approach it as a limit then isn't it the same basic idea?
I mean, I have an idea of the concept and some proprerties of R* but I'm not really familiarized with it, so I might be wrong, but.isn't n/oo actually defined as dx with a standard of 0 in it?
Farmageddon said:
|
Standard and non-standard calculus are equivalent in term of power. Some theorems are easier to prove in standard form, other are easier in non-standard; also, some people seem to like the more hands-on approach to infintesimal quantities that the non-standard calculus permits. In which you have the standard part as a "static" concept instead of the "limit" that is a sort of dynamic one - again, just preferences.
Digression:
In my experiences physicists already work with non-standard calculus in their head all the time, and they talk to each other in terms of dividing and multplying by infinitesimal quantities routinely. Non-standard calculus is just the clean formalization of that intuitive behaviour, no matter how many times your standard calculus teacher will try to hammer in your head that Integral(f(x)dx) is just a symbol and not the sum of products of f(x) by an infinitesimal quantity.
Digression over, anyway :)
You can work with limits or you can take the standard part of an R* number, and you get to the very same result ( I think that's called Leibnitz equivalence, actually ). But not even in non-standard calculus division by zero is contemplated, merely the division by an infinitesimal. So my point was that solving that equation (it has a solution in R*) is really a different issue from division by zero.
WereKitten said:
Standard and non-standard calculus are equivalent in term of power. Some theorems are easier to prove in standard form, other are easier in non-standard; also, some people seem to like the more hands-on approach to infintesimal quantities that the non-standard calculus permits. In which you have the standard part as a "static" concept instead of the "limit" that is a sort of dynamic one - again, just preferences. Digression: In my experiences physicists already work with non-standard calculus in their head all the time, and they talk to each other in terms of dividing and multplying by infinitesimal quantities routinely. Non-standard calculus is just the clean formalization of that intuitive behaviour, no matter how many times your standard calculus teacher will try to hammer in your head that Integral(f(x)dx) is just a symbol and not the sum of products of f(x) by an infinitesimal quantity. Digression over, anyway :) You can work with limits or you can take the standard part of an R* number, and you get to the very same result ( I think that's called Leibnitz equivalence, actually ). But not even in non-standard calculus division by zero is contemplated, merely the division by an infinitesimal. So my point was that solving that equation (it has a solution in R*) is really a different issue from division by zero. |
Ah, right.
I was actually thinking more in terms of comparing n/dx instead of actually n/0 to n^-oo. Reasoning being that working with R* would be the same as re-writing a limit in R as a number plus the infinitesimal (or infinite) part you can't really count in R.
So, in R*, does n^-oo actually equal 0 and not dx? It made me think, that means can you solve a Taylor series or somesuch for z = 0 in R*?
Sorry if I'm bothering, it made me curious :P
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