Wikipedia:Reference desk/Archives/Mathematics/2007 January 24

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January 24

Overbraces/underbraces

When to use underbraces and overbraces if there is a distinction? X [Mac Davis] (How's my driving?) 00:18, 24 January 2007 (UTC)

I think that's really a personal preference/stylistic question. I myself prefer underbraces. For instance, if I'm showing how to use the integration by parts formula by means of an example, here's what I might do. Since the formula is

$\int u\ dv=uv-\int v\ du,$

if I'm doing the example of $\int xe^{2x}\ dx$, where $u=x$ and $dv=e^{2x}\ dx$, then I'd write:

$\int \underbrace {x} _{u}\underbrace {e^{2x}\ dx} _{dv}=\underbrace {x} _{u}\cdot \underbrace {\left({\dfrac {1}{2}}\cdot e^{2x}\right)} _{v}-\int \underbrace {{\dfrac {1}{2}}e^{2x}} _{v}\underbrace {dx} _{du}$

instead of

$\int \overbrace {x} ^{u}\overbrace {e^{2x}\ dx} ^{dv}=\overbrace {x} ^{u}\cdot \overbrace {\left({\dfrac {1}{2}}\cdot e^{2x}\right)} ^{v}-\int \overbrace {{\dfrac {1}{2}}e^{2x}} ^{v}\overbrace {dx} ^{du}.$

I know this really isn't an answer, but I hope it helps somewhat. –King Bee ^{(T • C)} 03:16, 24 January 2007 (UTC)

Okay. "When it doesn't look funny" :P X [Mac Davis] (How's my driving?) 03:49, 24 January 2007 (UTC)

I tend to use overbraces if I'm indicating repetition:

$a^{b}=\overbrace {a\times a\times \cdots \times a} ^{\textstyle b\mathrm {\;factors} }$

—Bkell (talk) 05:22, 24 January 2007 (UTC)

steps to finding functions that satisfy certain rules

Hi

I need to find a function that satisfies these rules:

As x approaches 0, y approches a set limit. When x = 0, y = limit. x's limit >= y's limit

What steps are involved in finding this function?

Any help would be appreciated.

Thanks 203.196.50.55 05:31, 24 January 2007 (UTC)

Howdy, maybe it is just me, but I am not sure what you mean by "x's limit >= y's limit." What do you mean by the limit of x? --TeaDrinker 06:40, 24 January 2007 (UTC)

You probably want something like (x-1)/(x). Do I assume correctly? X [Mac Davis] (How's my driving?) 06:56, 24 January 2007 (UTC)

Not really what i am after

Say y's limit was 100, as x approaches 0, y would approach 100 to the point that when x = 0, y = 100. That's it. If you graphed it then it would look like the positive f(x) = 1/x, the problem with this function is x has to be > 0, otherwise the answer is a non-number not 100

Thanks again — Preceding unsigned comment added by 203.196.50.55 (talk)

How about

$y={\frac {x^{2}+x}{x}}$

where y approaches the limit 1 as x approaches 0, but, strictly speaking, cannot be evaluated at x=0.

Or how about

$y=x\sin({\frac {1}{x}})$

which approaches the limit 0 as x approaches 0, but cannot be evaluated at x=0 because 1/x is undefined there.

Is either of these close to what you are after ? Gandalf61 12:17, 24 January 2007 (UTC)

{ Note $y={\frac {x^{2}+x}{x}}$ = x+1 ? Gandalf?)87.102.10.13 13:54, 24 January 2007 (UTC)

Almost. ${\frac {x^{2}+x}{x}}$ and $x+1$ are actually different functions, which happen to take the same value for each value of x except when x is 0, when the first expression evaluates to 0/0, which is undefined. This is why I said strictly speaking above - it is a bit of a technical point. Gandalf61 14:07, 24 January 2007 (UTC)

mm, I agree that (x^2+x)/x cannot be evaluated 'as is' at x=0 but agree that the limit as x tends to zero tends to 1 , but think that the limit at x=0 is 1 and so can be evaluated at x=0 'as it stands' but by method of limits - the equations are the same I think...?87.102.10.13 14:22, 24 January 2007 (UTC)

I think the only problem here is that there are literally countless examples of what you have asked for - could you give more info on how the function approaches the limit at x=0 eg the shape/slope of the curve? Also do you mean that y is at it's maximum value when x=0.. does the value of y at x=infinity have to have a specific value..87.102.10.13 13:52, 24 January 2007 (UTC)

eg e^{-x^2}, 1/(1+x^2), 1+x^4/(3+x^8) etc.87.102.10.13 13:58, 24 January 2007 (UTC)

Do you mean that y must be undefined or imaginary when x<0? or should the graph be symmetrical about x=087.102.10.13 14:01, 24 January 2007 (UTC)

The first step in solving a problem is to have a clear enough idea of what the problem is, which no one has here. From the information given, we could propose a trivial solution.

${\begin{aligned}f\colon \mathbb {R} &{}\to \mathbb {R} \\x&{}\mapsto 100\end{aligned}}$

This clearly has the limit 100 as x approaches zero from either side! --KSmrq^T 17:04, 24 January 2007 (UTC)

Thanks for the multitude of impressive answers. Maybe a different scenario is what you need. I have an ideal number and two actual numbers. Say my ideal is 20 and my two actuals are 18 and 69, the difference between the ideal and the actual tells you how far away from the ideal the actual is, Now as the difference approaches 0, f(x) approaches its limit. f(x) has a range limit, say 0 - 100, where the difference (x) has no limit except 0. So f(x) = x + 100 will not work, zero is only approached from above and x will never be negative but will be 0 at which point f(x) will be 100.

Hope this scenario made it a bit clearer for you 203.196.50.55 00:19, 25 January 2007 (UTC)

It sounds like you mean y=100(1-x) (if the graph is a straight line - there still are other possibilities).?83.100.250.143 10:36, 25 January 2007 (UTC)

I have come up with a solution that satisfies my needs, just a simple function

f(x) = a^(x/b) * 100 where a and b are constants, a = 0.9 and b = 5 works well.

As x approaches zero, f(x) approaches 100 to the point that when x = 0, f(x) equals 100.

Thanks for all your help 203.196.50.55 11:22, 25 January 2007 (UTC)

type of geometry?

Is there a geometry where the sum of the cubes of lengths or magnitudes has useful meaning? eg c³=a³+b³?87.102.10.13 12:52, 24 January 2007 (UTC)

If you mean a geometry in which the equivalent of Pythagoras' theorem is a³+b³=c³, then the answer is yes. It is possible to define a geometry in which the distance between two points (x₁, y₁) and (x₂, y₂) is defined as

$d={\sqrt[{3}]{|x_{2}-x_{1}|^{3}+|y_{2}-y_{1}|^{3}}}.$

This would be called a L³ space - see the Motivation section of Lp space for more details. Gandalf61 13:59, 24 January 2007 (UTC)

Effectively yes I'm looking for something like that - but my motvation was this -the sum of powers of one eg c¹=a¹+b¹ is simply the addition of two colinear variables (needs only one dimension), c²=a²+b² is of course pythagorus and needs two dimensions (a plane), (pythagorus can be extended to three axises by taking one plane first - finding the length, then continuing with this length and the length of the third vector). But does c³=a³+b³ have no meaning in three dimensional euclidean space - if so this at least seems confusing to me...87.102.10.13 14:10, 24 January 2007 (UTC)

And does this mean that a one dimensional line is L¹ space and a two dimensional plane L² space, which makes me think that euclidean 3d space is just the combination of 2 L² spaces?87.102.10.13 14:15, 24 January 2007 (UTC)

Euclidean 3d space is an L² space, as is Euclidean 2d and 1d space. In each case the length of a vector (or distance of a point from the origin) is given by the root of the sum of the squars of the components, i.e.:

• 1-d space: d² = x₁²
• 2-d space: d² = x₁² + x₂²
• 3-d space: d² = x₁² + x₂² + x₃²

The sum c¹=a¹+b¹ is, well, neither here nor there. Yes, c = a + b works for adding two values together in 1 dimension. But c = a + b works for adding values in any number of dimensions - no powers needed. Rawling4851 14:59, 24 January 2007 (UTC)

If I add two vectors in 1d space the result is a1+a2 not sqrt(a1*a1+a2*a2), so I'd suggest that euclidean 1d space (a line) is L¹?87.102.10.13 15:03, 24 January 2007 (UTC)

And further on that euclidean 2d space (a plane) is L² space including/made up of two orthogonal L¹ spaces?? (that can be chosen in an infinite number of ways)87.102.10.13 15:07, 24 January 2007 (UTC)

If you look up the L^P space article, the p is not the power you use when adding vectors - it's the power you use when finding the length of a vector. When you add vectors together you don't use a power whatever the value of p. Rawling4851 15:08, 24 January 2007 (UTC)

Yes - that would be L¹ space when finding the length of the result of addition of vectors in 1d, (no power used because x¹ = x ??) eg 2-d space: d² = (x1+x2+x3...)² + (y1+y2+y3..)² 87.102.10.13 15:12, 24 January 2007 (UTC)

The p in an L^p space is the power involved in the definition of distance (which is called the space's metric). As Rawling4851 says, Euclidean geometry always takes place in an L² space, no matter how many dimensions you have, because the power involved in the definition of distances in Euclidean space is 2. Gandalf61 15:12, 24 January 2007 (UTC)

Right - I understood that - and I know that I can extend 'euclidean' space (L² space) up to higher dimensions mathematically (even if I can't visualise it). So I'm asking (seems more like philosophy that maths) why the space I think I'm inhabiting (it is at least the space I am perceiving at a given time) has 3 dimensions and a L² 'metric' - that is confusing to me - is there such a thing as an answer to this? There doesn't seem any particular reason why 3 dimensions with p=2 would be special/chosen over any other option..?87.102.10.13 15:20, 24 January 2007 (UTC)

I think that's just the way life is. Rawling4851 15:55, 24 January 2007 (UTC)

Thank you (it certainly is). If anyone knows any more please tell me though.87.102.10.13 15:58, 24 January 2007 (UTC)

Unless you live in Manhattan, which is in L¹ space (see Manhattan distance). --대조 | Talk 11:31, 31 January 2007 (UTC)

taylor series and infinite differentiability

Are there examples of infinitly differentiable functions that cannot be expressed as an (infinite) power series? eg f(x)=a₀x⁰+ a₁x¹+ a₂x²+ etc.. (or does an infinite power series encompass all? - is there a page that deals with this topic?)87.102.10.13 13:29, 24 January 2007 (UTC)

Every infinitely differentiable function has a Taylor series. However, some infinitely differentiable functions, like polynomials, will have a "finite" power series, since the (n + 1)^th derivative of a degree n polynomial is 0; hence every derivative thereafter is 0. Clicking the wikilink for Taylor series will provide you with more info. –King Bee ^{(T • C)} 13:43, 24 January 2007 (UTC)

Yes, (a finite power series is effectively an infinite power series with terms a_n=0 when n exceeds a given value).

I think what I was trying to say/ask was are all functions that are workable as a taylor series expressable as a power series?87.102.10.13 13:47, 24 January 2007 (UTC)

There are functions which are differentiable infinitely many times at a point and yet their Taylor expansion around that point (as well as any other power series) does not converge to the function anywhere but that point. For example, the function

$f(x)=\left\{{\begin{array}{ll}e^{\frac {-1}{x^{2}}}&x\neq 0\\0&x=0\end{array}}\right.$

Its Taylor expansion around 0 is the constant function 0, which isn't equal to the function in any positive radius.

This cannot happen for functions over the complex numbers, where (IIRC) it is enough for the function to be differentiable once for its Taylor series to converge to the function in some positive radius. -- Meni Rosenfeld (talk) 14:35, 24 January 2007 (UTC)

Thanks - is it right to say that the differential of the function (2x^-3e^{x^-2} is undefined at x=0) (although it appears to be 0 it changes sign crossing x=0?). This would be a reason why it (taylor) doesn't work?87.102.10.13 14:50, 24 January 2007 (UTC)

So effectively there is a discontinuity at x=0 in the first differential (even though the magnitude of the discontinuity is 0?)?87.102.10.13 14:52, 24 January 2007 (UTC)

Not exactly. First, the word you were looking for is derivative, not differential, which is a different thing. Second, it is true that for any x≠0, $f'(x)={\frac {2}{x^{3}}}e^{\frac {-1}{x^{2}}}$, but this formula doesn't apply to x=0. For this case, you will need to return to the limit definition, and you will get f'(0)=0. In other words, we have

$f'(x)=\left\{{\begin{array}{ll}{\frac {2}{x^{3}}}e^{\frac {-1}{x^{2}}}&x\neq 0\\0&x=0\end{array}}\right.$

Which is a continuous function. Not only is it continuous, it is differentiable (everywhere), and you can repeat the process. The original function therefore has infinitely many derivatives, from which it follows that they are all continuous. But they are all equal to 0 at x=0, from which it follows that the Taylor expansion around that point is 0. Intuitively, this function is extremely close to 0 near x=0 - closer to 0 than any power of x.

This function becomes very wild when you consider complex values for x, in which case it will not even be continuous at 0. Since a Taylor expansion, when it works, should also work for complex values, and it cannot work here for complex values, it does not work at all. -- Meni Rosenfeld (talk) 15:07, 24 January 2007 (UTC)

Yes thanks - I couldn't help noticing that the expansion of the original function gives terms 1- 1/x^2 + 1/2x^4 - etc - the individual terms of which all are discontinuous at x=0 (unlike a power series) - it's clear that the function plots continuously at x=0 despite this - but I still wonder if 'technically'(for want of a better word) there is a sort of discontinuity at x=0 even if this discontinuity results in no net change in value??87.102.10.13 15:35, 24 January 2007 (UTC)

OK thanks, I've understood it now - the function is composed of a sum of powers of the type x^-n which individually work as taylor expansions provided they aren't expanded about x=0 were the function is infinite. So the situation continues in a sum of these functions (the taylor series can be summed too), so despite the fact that e^{-x^-2} appears continuous at x=0 it is constructed from a series of functions that are at discontinuous at x=0, so the expansion continues not to work at x=0.87.102.10.13 15:56, 24 January 2007 (UTC)

See also Non-analytic smooth function. – b_jonas 18:55, 24 January 2007 (UTC)

Although I didn't get an answer to my original question (I've found the answer in the meantime) thanks to you all who responded - I've learnt a little about something else instead. Thank you.87.102.10.13 21:18, 24 January 2007 (UTC)

square feet

this may seem the ultimate dumb question, but ive nevr mastered this basic skill. with the answer, perhaps i'll know it from now on. a storage shed i'm looking at renting is discribed as 122 square feet.- so, in feet, what is the width and lenght of the room?-----------also, does 'square feet' and 'feet square'mean the same thing? —The preceding unsigned comment was added by 66.244.68.18 (talk) 15:10, 24 January 2007 (UTC).

122 square feet is the area, in square feet, so it might be 2 feet wide and 61 feet long, say. If something was 122 feet square, on the other hand, it would be 122 feet wide and 122 feet long. Rawling4851 15:32, 24 January 2007 (UTC)

Heh, "feet" is a funny word. Rawling4851 15:32, 24 January 2007 (UTC)

It could be approx 11 by 11 feet (121) that should give you an impression of the size of it.87.102.10.13 15:37, 24 January 2007 (UTC)

Read the description carefully. One common shed is 8 feet by 3 feet, said to have a storage area of 23 square feet and 122 cubic feet. So it is the volume, not the area, that is 122 for this shed. If you hold your arms straight out to the sides, the distance from fingertip to fingertip is approximately your height; this would be a rather narrow shed, with barely room inside to turn around. --KSmrq^T 19:43, 24 January 2007 (UTC)

how to?

How couĆld i draw a triangle that has three bisectors that are 1", 1"1/4 and 1"1/2 —The preceding unsigned comment was added by 68.209.69.103 (talk) 23:41, 24 January 2007 (UTC).

"Bisector" is ambiguous. The lengths specified could be (any of these, at least) (a) of the medians, i.e. the lines joining each vertex to the midpoint of the opposite side; (b) of the angle bisectors, i.e. the lines joining each vertex to the incentre; (c) of the side bisectors, i.e. the lines from the midpoint of each side to the circumcentre. I assume that in each case the length of the sides of the triangle could be calculated from the given values, so which did you mean?86.132.235.173 14:39, 25 January 2007 (UTC)