# Wikipedia:Reference desk archive/Mathematics/2006 August 20

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## Italian Mathematican / astronomer

Hi, I am interested in finding about an Italian mathematican, maybe 15th century, who later was also a renowned astronomer. Seems he predicted his own death and with little time remaining and he being still alive, he hanged himself! Does anyone know anything? Thanks, 82.190.241.130 05:40, 20 August 2006 (UTC)

Check out Gerolamo Cardano and the corresponding talk page.Sluzzelin 06:38, 20 August 2006 (UTC)

Thanks, its Sig. Cardano I was looking for. Have a nice time.......Neelzaff 07:10, 20 August 2006 (UTC)

Strange...I was looking for the same thing just now...Thanks for asking and saving me the work!02:06, 23 August 2006 (UTC)

## Interesting Mushroom

I was playing golf in my backyard once and I hit a mushroom with the club, it looked white inside and common enough so I didn't pay it much attention, but when I came back 5 minutes later it was bright blue. I picked it up and snapped it and watched it turn from white to Prussian blue in about 20 seconds before my eyes.

The location was the Southern Highlands, NSW, Australia.

Does anyone know what kind of mushy it was? What chemicals are involved and how toxic is it? --WB Frontier 12:26, 20 August 2006 (UTC)

PS, I'm not at my computer very often, if you know the answer please also write it in my user page. Cheers.

It does sound pretty cool, but this belongs on the Science Desk, not Math. StuRat 14:30, 20 August 2006 (UTC)
Since I'm too lazy to move this to the sci desk, I'll just answer. The best match I've found is "Gyroporus cyanescens", aka the "blue-staining bolete" mushroom. There is a nice website about the mushroom complete with a video of the staining here. A google search for "Gyroporus cyanescens" yields several other sites. I haven't seen any sites specifically mention Australia as a habitat, but it seems to be widespread throughout N America and Europe. Hope it's what you're looking for! --Bmk 16:27, 20 August 2006 (UTC)
And btw, according to the cited site (how awkward), the chemical reaction is variegatic acid + oxygen + enzyme ==> quinone methide of variegatic acid. I'm no chemist, but that's what the site says. And it seems to say that it is good to eat, but for god's sake, don't eat it!!! It's very easy to misidentify mushrooms, and many of them are fatal to humans. Just don't eat it. It isn't worth the risk. --Bmk 17:16, 20 August 2006 (UTC)
Many Psilocybe species also stain blue, so you may have found yourself some magic mushrooms. —Keenan Pepper 06:22, 22 August 2006 (UTC)

## Significant Figures

If I'm dividing 2 numbers, 12442.23/2.2, then the answer must have only 2 significant figures. That's not possible right? So could you please explain that.

I also don't understand if I divide 1222/2, then it would mean 611, right, because they have infinite significant figures.

What you need to do is this: if the number 12442.23 is known to be accurate to 2 decimal places, then it is known to be between 12442.225 and 12442.235. If 2.2 is known to be accurate to 1 decimal place, then it is known to be between 2.15 and 2.25. Then the result of the division must be between 12442.225/2.25 and 12442.235/2.15. That way you get precise limits for the quotient - better than rough guides like " ... must have only 2 significant figures ... ". Madmath789 18:30, 20 August 2006 (UTC)
thanks a lot, but we are required to use this sig figs thing.
Aha! - so it is homework after all? :-) Madmath789 19:27, 20 August 2006 (UTC)
Using the incredibly inaccurate "significant figures" method, the 2.2 has only two significant figures, so the answer is limited by that, to two significant figures. That means your answer is 5700 instead of 5656. StuRat 19:08, 20 August 2006 (UTC)
Let's switch to interval arithmetic, eh? (Or not.) Meanwhile, two thoughts: Are we talking about significant places after the decimal point or significant figures? Also, let's not lose sight of the big picture, which is that it is a Good Thing to be aware that most practical computations neither begin nor end with exact values. --KSmrqT 05:39, 21 August 2006 (UTC)
So, would 2.00*10^2 be one or 2 sig figs.
I would consider 2*10^2 to have 1 sig. fig., 2.0*10^2 to have 2 sig.fig. and 2.00*10^2 to have 3 sig.fig. - but others may differ on this. Madmath789 20:28, 20 August 2006 (UTC)
It should be noted that the result has the same accuracy as the least accurate thing that goes in only is a rule of thumb, it's not universally true. The result can be both more and less accurate than what goes in. Look for example at this square wave. If we put in a value of ${\displaystyle t}$ that wiggles between slightly smaller than 1 to slightly greater than 1, the change in ${\displaystyle f(t)}$ is dramatic. On the other hand, if ${\displaystyle t}$ wiggles around the value 2, ${\displaystyle f(t)}$ is very steady. —Bromskloss 00:14, 21 August 2006 (UTC)
I agree, significant figures are a highly inaccurate way to deal with error, and I wish they would stop teaching that method in schools. I much prefer + and - values after each number. Those quite accurately portray the range of possible values, but don't show the distribution within that range. For that, you need to get into statistics notation. StuRat 01:09, 21 August 2006 (UTC)
Actually, significant figures is not the same thing as the rule of thumb (what goes in, comes out) I referred to. Significant figures is actually a special case of the ± notation you mentioned. Let me explain. Suppose we have the number 1000 with 4 significant figures. That is the same as writing "1000 ± 0,5". The same number with only 3 significant figures is "1000 ± 5", and with 2 it becomes "1000 ± 50". As we can see, it is a special case because only certain numbers can appear after the "±" sign, otherwise we would have to talk about fractions of significant figures. Note also that the possible values after "±" depend on which base we choose. —Bromskloss 07:17, 21 August 2006 (UTC)

## Quotient rule

I don't understand the quotient rule as it relates to at least two problems. Help me out! --Neutralitytalk 19:28, 20 August 2006 (UTC)

• Find ${\displaystyle d^{2}y/dx^{2}}$ given that ${\displaystyle y=(4-x)/x}$
• Find the equation of the tangent to ${\displaystyle y=f(x)}$ for ${\displaystyle f(x)}$ equal to ${\displaystyle (x^{2}+4)/x)}$ at ${\displaystyle x=-1}$

${\displaystyle d^{2}y/dx^{2}}$ is equal to the second derivative ${\displaystyle (f''(x))}$, right? Neutralitytalk 19:30, 20 August 2006 (UTC)

Yes, that is indeed the second derivative. The quotient rule isn't actually needed for any of these problems (although you could use it). You haven't explained what are you having trouble with - they're both very straightforward problems. -- Meni Rosenfeld (talk) 19:51, 20 August 2006 (UTC)
How would you do this without the quotient rule? I suppose you could just say that y = 4/x - 1, and differentiate that? --HappyCamper 19:53, 20 August 2006 (UTC)
The quotient rule is an elaboration of the product rule. You could indeed split the fraction as you mentioned, or you could treat it as a product; ${\displaystyle (4-x)x^{-1}}$ --Bmk 20:06, 20 August 2006 (UTC)
Yes, HappyCamper's suggestion is what I had in mind. -- Meni Rosenfeld (talk) 20:16, 20 August 2006 (UTC)
Can somebody work the problem? I still don't understand how one would get y = 4/x - 1. _Sorry, I'm a social sciences/humanities person.) ;) Neutralitytalk 21:29, 20 August 2006 (UTC)
We all can work the problems, but how will it help you if we solve them? Understanding general principles is by far more important than any specific example or problem. Laws of algebra are a start; for example, the one stating that
${\displaystyle {\frac {a-b}{c}}={\frac {a}{c}}-{\frac {b}{c}}}$
Once you've mastered elementary algebra (your last question implies that this is not yet the case), you can move on to calculus, where there are some rules for finding derivatives (which are all less important than understanding what a derivative is). The ones you need here are:
${\displaystyle (x^{n})'=nx^{n-1}}$
${\displaystyle (f+g)'=f'+g'}$
${\displaystyle (af)'=af'}$
This, and the idea that a second derivative is the derivative of the derivative, is all you need to know to solve the first problem. The second problem involves some more elementary algebra. We'll be happy to help if there's any specific issue you're having trouble understanding. -- Meni Rosenfeld (talk) 22:23, 20 August 2006 (UTC)
One of the magic keys to learning is coping with a mystery. Presented with a problem to solve and no obvious solution, how do we make forward progress? A fruitful strategy is to get specific. What exactly do we know? What exactly do we want? Can we pin down a narrow question? It has often been observed that asking a good question is the essential step to progress in mathematics, and science, and even business. Recommended reading is George Pólya's How to Solve It.
So, let's look at your example. You began with an excellent question: "Does d2y/dx2 mean the second derivative of y with respect to x?" That isolated a critical uncertainty. And you have also begun to focus on the quotient rule as relevant to both (4−x)/x and (x2+4)/x. Now you need to keep generating questions.
The quotient rule gives a prescription for the derivative of a ratio of two functions. So you need to fit the rule to your data. In each case, what is the numerator function, what is the denominator function, what are their derivatives, and how does the quotient rule bring all that together? And in the first problem, how can the quotient rule — which only gives a first derivative — be used to find a second derivative? And so on.
This kind of "pattern matching" and assembling partial solutions into a complete path is an essential skill in all problem solving. That's just as true in the social sciences and humanities as it is in mathematics and the physical sciences. If anything, it's easier in mathematics. Compare this simple exercise to a problem in law, where there are myriad conflicting statutes and precedents, and the task of applying knowledge of rhetorical techniques to convincing a judge or jury or negotiation opponent to see things your way! (And there is no "right" answer.) --KSmrqT 06:20, 21 August 2006 (UTC)

## Linearity

The article on linearity states, quite correctly, that the conditions

${\displaystyle f(x+y)=f(x)+f(y)}$,
${\displaystyle f(ax)=af(x)}$

are equivalent to

${\displaystyle f\left(\sum _{i=1}^{m}a_{i}x_{i}\right)=\sum _{i=1}^{m}a_{i}f\left(x_{i}\right)}$.

Note that the sums have a finite number of terms. Could someone construct an example (with ${\displaystyle f}$ of type ${\displaystyle \mathbb {R} \to \mathbb {R} }$) where

${\displaystyle f\left(\sum _{i=1}^{\infty }a_{i}x_{i}\right)=\sum _{i=1}^{\infty }a_{i}f\left(x_{i}\right)}$

(infinite number of terms) does not hold? Actually, I was about to be more general and use a sum of type ${\displaystyle {\begin{matrix}\sum _{i\in S}\end{matrix}}}$, where ${\displaystyle S}$ may be an uncountable set, but is such a sum (with so many terms) even defined? —Bromskloss 22:31, 20 August 2006 (UTC)

Well, yes, I suppose you could go for something as simple as let f be the zero function, and let the ${\displaystyle a_{i}}$ and ${\displaystyle x_{i}}$ be such that their sum does not exist (eg. ${\displaystyle a_{i}=1}$ and ${\displaystyle x_{i}=i}$). Then the LHS would technically not exist (with a bit of work you could probably construct a function that really can't be defined all the way out to infinity), while the RHS is just zero. As for the sum of an uncountable set, I believe it can only be defined for the case when ${\displaystyle x_{i}}$ is zero almost everywhere. 129.78.64.105Confusing Manifestation 00:33, 21 August 2006 (UTC)
OK, you found a loophole – the sums might not converge. So let me instead ask for an example where they do. —Bromskloss 00:50, 21 August 2006 (UTC)
What does almost everywhere mean in this case? Could it be that the points that remain after almost everywhere has been removed actually are countable? I don't really know much about measure theory, so mabye you are referring to some natural measure that makes this obvious. —Bromskloss 00:50, 21 August 2006 (UTC)
'Almost everywhere' means for all x except in a set of measure 0 (see, probably measure theory or Lebesgue measure or even almost everywhere, I think). Essentially, yes, it means except for a countable set. Confusing Manifestation 12:10, 21 August 2006 (UTC)
If f is of type ${\displaystyle \mathbb {R} \to \mathbb {R} }$ and linear, then the property "${\displaystyle f(ax)=af(x)}$ for all aR" implies that ${\displaystyle f(x)=xf(1)}$. Hence
${\displaystyle f\left(\sum _{i=1}^{\infty }a_{i}x_{i}\right)=\sum _{i=1}^{\infty }a_{i}x_{i}f(1)=\sum _{i=1}^{\infty }a_{i}f\left(x_{i}\right),}$
assuming that all sums converge. I think you need the domain of f to be infinite-dimensional to get a counterexample. One possibility is to consider Q-linear functions from R to R, which are functions f satisfying
• ${\displaystyle f(x+y)=f(x)+f(y)}$ for all x, yR,
• ${\displaystyle f(ax)=af(x)}$ for all xR, and aQ.
An example of a Q-linear function is the function f with f(x) = 0 if x is a rational number and f(x) = 1 if x is irrational (this does not quite work, see below). For this function,
${\displaystyle f\left(\sum _{n=0}^{\infty }{\frac {1}{n!}}\right)=f(e)=1}$
and
${\displaystyle \sum _{n=0}^{\infty }f\left({\frac {1}{n!}}\right)=\sum _{n=0}^{\infty }0=0}$
are not equal. I suppose there is a nicer example, but I can't find it at the moment.
For the sum over an uncountable index set to be defined, you need that the set of indices i for which ai is nonzero, is finite (I meant "countable", see Confusing Manifestation's comment below) -- Jitse Niesen (talk) 04:35, 21 August 2006 (UTC), amended 02:37, 22 August 2006 (UTC)
Actually, you just need countable, not finite, nonzero terms, as long as you then get an absolutely convergent series. Confusing Manifestation 12:17, 21 August 2006 (UTC)
Jitse, your example does not satisfy Q-linearity. Perhaps ${\displaystyle f(x)=0}$ if ${\displaystyle x}$ is rational and ${\displaystyle f(x)=x}$ otherwise? (Note added: actually, not even that works, because ${\displaystyle f({\sqrt {2}})\neq f({\sqrt {2}}-1)+f(1)}$. Perhaps you need to go to infinite dimensional spaces to make this work. For example, you could consider the space of functions on [0,1] with finite left and right hand limits and let ${\displaystyle f}$ be the sum of the function's jump discontinuities. Then there exist sequences of smooth functions which sum to discontinuous ones, say the Fourier series of the Heaviside step function.) –Joke 13:52, 21 August 2006 (UTC)
Ouch, that's two big mistakes. I need to be more careful with my example. Unfortunately, that makes it quite ugly. The way to go, I think, is to extend {1,e} to a Q-Hamel basis of R, define f(e) = 1 and f(x) = 0 for all other elements of the basis, and extend by linearity. I think Joke's example is better. -- Jitse Niesen (talk) 02:37, 22 August 2006 (UTC)

Maybe there's something wrong with this argument, but I think we can make it work as long as we assume that the function is both linear and continuous, Then define ${\displaystyle S_{N}=\sum _{i=1}^{N}a_{i}x_{i}}$, and we have ${\displaystyle f(\sum _{i=1}^{\infty }a_{i}x_{i})=f(\lim _{N\to \infty }S_{N})=\lim _{N\to \infty }f(S_{N})=\lim _{N\to \infty }f(\sum _{i=1}^{N}a_{i}x_{i})=\lim _{N\to \infty }\sum _{i=1}^{N}a_{i}f(x_{i})=\sum _{i=1}^{\infty }a_{i}f(x_{i}).}$ I don't think we need absolute convergence, although clearly if the sums are not absolutely convergent each depends on ordering. Moreover, since all ${\displaystyle \mathbb {R} }$-linear functions in finite dimensions are continuous, I think this gives it to us. -- Deville (Talk) 15:05, 21 August 2006 (UTC)