Wikipedia:Reference desk/Archives/Mathematics/2007 December 16

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December 16

Dissimilar Symbol

Hello. Is there such thing as a dissimilar symbol (e.g. maybe a slashed tilde) in comparing similar triangles? If so, like how does it look? Thanks in advance. --Mayfare (talk) 19:16, 16 December 2007 (UTC)

Pretty much any mathematical relation can be negated with a slash. In the case of similarity that would be $\not \sim$. I don't know if it is widely used, though. -- Meni Rosenfeld (talk) 19:24, 16 December 2007 (UTC)

Inverse functions

Does the function y=e^x-x have an inverse? How can you prove it doesn't if it doesn't? The reason for this question is in that I'm interested in that e^pi-pi is very close to 20, and I'm wondering what the reason for this is. Indeed123 (talk) 19:23, 16 December 2007 (UTC)

The function does not have an inverse, since it is not one-to one; for every $y>1$ there are two values of x such that $e^{x}-x=y$. However, you can find an inverse if you restrict the domain of x - for any $y\geq 1$ you can define $g(y)$ to be the unique x such that $x\geq 0$ and $e^{x}-x=y$. Unfotruantely, this function g cannot be expressed with the elementary functions found on a typical calculator. If it helps in any way, you have $g(20)=3.141633$, which for some reason (probably a coincidence) is close to π. -- Meni Rosenfeld (talk) 19:35, 16 December 2007 (UTC)

Just what I was about to say! The article Mathematical coincidence has this example and many others. AndrewWTaylor (talk) 19:39, 16 December 2007 (UTC)

The solution with x ≤ 0 can be expressed in closed form if you are allowed to use the (non-elementary) Lambert W function:

$x=-y-W(-e^{-y})\,.$

No coincidences that side of zero that I'm aware of, I'm afraid.  --Lambiam 00:05, 17 December 2007 (UTC)

I guess most CAS have a generalized version of the Lambert function, $W_{i}(\bullet )$giving the i^th solution branch.

With that in mind, you have for x > 0:

$x=-y-Re(W_{i}(-e^{-y}))$

i being any positive integer. Pallida  Mors 18:42, 17 December 2007 (UTC)

1/0=Infinity

Infinity can never be defined, therefore shouldn't it be equal to 1/0. 1/0 is greater than any imaginable number, as well as infinity. 1/0 being greater than any other number can be shown if you start taking any real number and plugging it in for the zero. When this happen the answer to the algebraic equation gets greater as the denominator gets lower. This brings me to conclude that 0 when plugged into 1/x would be the largest number, infinity. Even in the graph 1/x, it can been seen that when x is equal to zero, it crosses the y-axis at a point that is undefined, though logically infinity. Therefore, through this equation is possible to postulate that infinity and negative infinity are one and the same. The reason for this is since the equation is a function, meaning it crosses the y-axis at one point, it cannot have two y-axis crossings, unless infinity and negative infinity are one and the same. —Preceding unsigned comment added by ARedens (talk • contribs) 23:35, 16 December 2007 (UTC)

If you read our article on Infinity, you will see that there are several different notions in mathematics that may be called in some sense "infinite" or "infinity", and that have quite precise definitions. Usually, in mathematics, to be able to say that two things are equal requires that both are defined. 1/0 is not defined (in the usual interpretation of the symbols in this formula). The limit of sin x as x goes to infinity is also not defined. You can't say then that that limit is equal to 1/0.  --Lambiam 23:48, 16 December 2007 (UTC)

If you are talking about $\infty$ as I think you are, then yes it can actually be defined, but the VALUE of $\infty$ cannot, thus infinity is a concept not a number. There obviously can't be a "largest number" since you could take that number "x" and add 1 and have a "bigger" number. What we can conclude about 1/x at x=0 is that the function has a non-removable discontinuity there, that as x approaches 0 from negative values 1/x approaches negative infinity and as x approaches 0 from positive values 1/x approaches positive infinity. If you have access to a graphing utility try graphing y=1/x and see what it looks like, that should clarify anything that you may be uncertain about. A math-wiki (talk) 00:37, 17 December 2007 (UTC)

Not quite. Infinity is first and foremost a concept which can have different interpretations in different contexts, but as Lambiam said, in many structures you can take some formally constructed object and call it "infinity" or denote it by ∞, and you can say that some objects have the property of being "infinite". Whether you choose to call such objects "numbers" or not is just a matter of terminology. My favorite example is the real projective line, where a "number" denoted by ∞ exists and satisfies 1/0 = ∞. Another is the extended real number line, in which there is a largest number, $+\infty$. Your claim $x<x+1$ is true for real numbers but there is no reason why that should hold for other structures. I have no idea what you mean by "the value of ∞".

To the OP: Your very last statement is absolutely correct. 1/0 = ∞ would mean that positive and negative infinity are the same - this is exactly what is done in the real projective line. -- Meni Rosenfeld (talk) 09:36, 17 December 2007 (UTC)

If you are asking if 1/0 = $\infty$ in the realm of real numbers, the answer is no. 1/0 = undefined. a * b = c, c / a = b. Subsutiting #s for a = 0, b = 1, c = 0. 0 * 1 = 0, 0 / 0 = 1. Substing a = 0, b = 2, c = 0. 0 * 2 = 0, 0 / 0 = 2. So we have 0 / 0 = 1 and 2 at the same time. n / 0 = undefined. n / m as m approaches 0 = $\infty$. 2/m is always 2x greater than 1/m but 2/m is not always greater than 1/o as m and o approaches 0. Whether 2/m is greater than 1/o depends on the rate m and o approaches 0. So infinity is defined but is not a number. Infinity is larger then all real numbers but since infinity is not a number, it is not possible to determine the largest infinity. NYCDA (talk) 18:59, 17 December 2007 (UTC)

No offense, but can we please stick to coherent, informed responses? Indeed, 1/0 is undefined in the real numbers, but "undefined" is a metamathematical description rather than a mathematical object, so you cannot write "1/0 = undefined". There is no universally accepted definition of "number", thus the statement "Infinity is not a number" is meaningless. "Infinity is not a real number" is true, though. The kind of infinity that interest us here is unsigned, and is thus not larger than anything. -- Meni Rosenfeld (talk) 19:08, 17 December 2007 (UTC)

I guess I'm having a hard time understanding what you mean by cannot write "1/0=undefined". If on the test the question was solve 1/0, I would write undefined. If asked if infinity is the largest number there is, I would answer no since infinity + 1 = infinity but also infinity + 1 > infinity. Is 2/0 greater than 1/0? Maybe. But 2/n is always greater than 1/n even if n = 0. n/n = 1 even if n = 0 but 0/0 is undefined. NYCDA (talk) 23:05, 21 December 2007 (UTC)

Sorry, NYCDA, but you don't know what you're talking about."But 2/n is always greater than 1/n even if n = 0." Is this true for n = -1/2? "Is 2/0 greater than 1/0? Maybe." That's a meaningless question. How exactly can you solve 1/0?

Differential equation

I was recently introduced to the differential equation question, "Solve the differential equation $(x^{2}+1){\frac {dy}{dx}}-xy=0$ where $x>0$, $y>0$ given that $y=1$ when $x=1$." To list out all my steps here would take some time, but to summarize: I first moved xy to the right side, then expanded the left side, then divided all terms by x and then finally divided all terms by dy, to end up with:

$x{\frac {1}{dx}}+{\frac {1}{x}}{\frac {1}{dx}}=y{\frac {1}{dy}}$

Integrating away the dxes and dy,

${\begin{aligned}\int {\frac {1}{x}}dx+\int xdx&=\int {\frac {1}{y}}dy\\\ln {x}+{\frac {1}{2}}x^{2}+C&=\ln {y}\\\end{aligned}}$

Knowing that (1,1) was a point on the curve I determined C to be -0.5 and ended up with the function

$y=e^{\ln {x}+{\frac {1}{2}}x^{2}-{\frac {1}{2}}}$

Problem is, if I put this function in my graphing calculator, it determines ${\frac {dy}{dx}}$ to be 2 at (1,1), however it should be 0.5 according to the intial question ${\Bigg (}{\frac {dy}{dx}}={\frac {xy}{x^{2}+1}}{\Bigg )}$. I can't see what I've done wrong, but I doubt the calculator has made a mistake, so, what have I done wrong? Thanks --Colonel Cow (talk) 23:57, 16 December 2007 (UTC)

You have gone from $x{\frac {1}{dx}}+{\frac {1}{x}}{\frac {1}{dx}}=y{\frac {1}{dy}}$ to ${\frac {1}{x}}dx+xdx={\frac {1}{y}}dy$, which is invalid since ${\frac {1}{a+b}}\neq {\frac {1}{a}}+{\frac {1}{b}}$. In as much as this kind of thing makes sense at all, you want $\int {\frac {1}{x+{\frac {1}{x}}}}dx=\int {\frac {1}{y}}dy$, which gives the correct answer. Algebraist 00:26, 17 December 2007 (UTC)

I think your error is here: $x{\frac {1}{dx}}+{\frac {1}{x}}{\frac {1}{dx}}=y{\frac {1}{dy}}$

That much is true, but you then take the reciprocal of each term individually, which is wrong. Think of it this way $(a+b)^{-1}\neq a^{-1}+b^{-1}$. Instead, you should have the equation: $\int {\dfrac {dy}{y}}=\int {\dfrac {x}{x^{2}+1}}dx$, which, i have to say, is not especially pretty, since x² + 1 is irreducible in R by partial fractions. mattbuck (talk) 00:21, 17 December 2007 (UTC)

(after edit conflict) The problem is in the step in which you introduce the integral signs. You seem to be doing something like taking the inverse of a sum by taking a sum of inverses. To solve this, you should try to achieve separation of variables: all x's on one side of the equation and all y's on the other side.  --Lambiam 00:24, 17 December 2007 (UTC)

Partial Fractions is not the way to solve what you have there mattbuck (in fact the way is easier than partial fractions), the trick to the right side is to notice that the numerator is the 2 times derivative of the denominator. So mulitply and divide by 2 and pull out 1/2 outside the integral, then use the integral of du/u to solve it. OP: You should also note that you got 2 when you should've got .5, a subtle clue that your mistake was inverting the way you did. Lambian: He separated his variables ok, it was what he did next that was in error. A math-wiki (talk) 00:33, 17 December 2007 (UTC)

That's a good observation you make A math-wiki about the 2/0.5.

Anyway, you guys combined with a friend of mine have shown me the better way to do this question. However, I don't understand EXACTLY where I made the issue. I mean, from what I know

$\int x{\frac {1}{dx}}=\int {\frac {1}{x}}dx$

at least, on its own. But now it appears that isn't true, so what's wrong with doing that? Thanks --Colonel Cow (talk) 00:42, 17 December 2007 (UTC)

Well, nevermind -- the more I look at that, the more I realise how ridiculous that is (I mean, they're inverses, x/dx vs dx/x).

Thanks guys for your help. Just for the record, the correct answer was indeed $y={\Bigg (}{\frac {x^{2}+1}{2}}{\Bigg )}^{\frac {1}{2}}$. --Colonel Cow (talk) 00:58, 17 December 2007 (UTC)

You should drop the 1/2 since it's a constant $y={\Bigg (}{x^{2}+1}{\Bigg )}^{\frac {1}{2}}$. NYCDA (talk) 23:47, 21 December 2007 (UTC)

No, you shouldn't drop it.