# Man arrested at Large Hadron Collider claims he's from the future

A would-be saboteur arrested today at the Large Hadron Collider in Switzerland made the bizarre claim that he was from the future. Eloi Cole, a strangely dressed young man, said that he had travelled back in time to prevent the LHC from destroying the world.

The LHC successfully collided particles at record force earlier this week, a milestone Mr Cole was attempting to disrupt by stopping supplies of Mountain Dew to the experiment's vending machines. He also claimed responsibility for the infamous baguette sabotage in November last year.

Mr Cole was seized by Swiss police after CERN security guards spotted him rooting around in bins. He explained that he was looking for fuel for his 'time machine power unit', a device that resembled a kitchen blender.

Police said Mr Cole, who was wearing a bow tie and rather too much tweed for his age, would not reveal his country of origin. "Countries do not exist where I am from. The discovery of the Higgs boson led to limitless power, the elimination of poverty and Kit-Kats for everyone. It is a communist chocolate hellhole and I'm here to stop it ever happening."

This isn't the first time time-travel has been blamed for mishaps at the LHC. Last year, the Japanese physicist Masao Ninomiya and Danish string-theory pioneer Holger Bech Nielsen put forward the hypothesis that the Higgs boson was so "abhorrent" that it somehow caused a ripple in time that prevented its own discovery.

Professor Brian Cox, a CERN physicist and full-time rock'n'roll TV scientist, was sympathetic to Mr Cole. "Bless him, he sounds harmless enough. At least he didn't mention bloody black holes."

Mr Cole was taken to a secure mental health facility in Geneva but later disappeared from his cell. Police are baffled, but not that bothered.

# Women are worse than men at turning networks to their advantage

Old boy in the chair

IN THE rarefied world of the corporate board, a good network matters. Recruitment often involves word-of-mouth recommendations: getting on a shortlist is easier if you have the right connections. New research suggests men use contacts better than women.

Marie Lalanne and Paul Seabright of the Toulouse School of Economics measure the effect of a network on remuneration using a database of board members in Europe and America. They find that if you were to compare two executive directors, identical in every way except that one had 200 ex-colleagues now sitting on boards and the other 400, the latter, on average, would be paid 6% more. For non-executives the gap is 14%.

The really juicy finding concerns the difference between the sexes. Among executive-board members, women earn 17% less than their male counterparts. There are plenty of plausible explanations for this disparity, from interruptions to women’s careers to old-fashioned discrimination. But the authors find that this pay gap can be fully explained by the effect of executives’ networks. Men can leverage a large network into more senior positions or a seat on a more lucrative board; women don’t seem to be able to.

Women could just have weaker connections with members of their networks. “Women seem more inclined to build and rely on only a few strong relationships,” says Mr Seabright. Men are better at developing passing acquaintances into a network, and better at maintaining a high personal profile through these contacts. Women may, of course, also be hurt by the existing dominance of men on boards and a male preference for filling executive positions with other men. But a tendency to think of other men first will be amplified if talented women don’t stay on the radar.

Interestingly, there is only a marginal pay difference between men and women when it comes to non-executive directors, and no difference in the effectiveness of their networks. It is possible that this reflects pressure for “gender quotas” on corporate boards. Women are able to find their way onto shortlists for lower-paid, non-executive positions. But that’s not where the real power lies.

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# Q: What does 0^0 (zero raised to the zeroth power) equal? Why do mathematicians and high school teachers disagree?

Clever student:

I know!

$x^{0}$ =  $x^{1-1}$ = $x^{1} x^{-1}$ = $\frac{x}{x}$ = $1$.

Now we just plug in x=0, and we see that zero to the zero is one!

Cleverer student:

No, you’re wrong! You’re not allowed to divide by zero, which you did in the last step. This is how to do it:

$0^{x}$$0^{1+x-1}$$0^{1} \times 0^{x-1}$$0 \times 0^{x-1}$$0$

which is true since anything times 0 is 0. That means that

$0^{0}$ = $0$.

Cleverest student :

That doesn’t work either, because if $x=0$ then

$0^{x-1}$ is $0^{-1} = \frac{1}{0}$

so your third step also involves dividing by zero which isn’t allowed! Instead, we can think about the function $x^{x}$ and see what happens as x>0 gets small. We have:

$\lim_{x \to 0^{+}} x^{x}$ = $\lim_{x \to 0^{+}} \exp(\log(x^{x}))$

= $\lim_{x \to 0^{+}} \exp(x \log(x))$

= $\exp( \lim_{x \to 0^{+} } x \log(x) )$

= $\exp( \lim_{x \to 0^{+} } \frac{\log(x)}{ x^{-1} } )$

= $\exp( \lim_{x \to 0^{+} } \frac{ \frac{d}{dx} \log(x) }{ \frac{d}{dx} x^{-1} } )$

= $\exp( \lim_{x \to 0^{+} } \frac{x^{-1}}{- x^{-2}} )$

= $\exp( \lim_{x \to 0^{+} } -x )$

= $\exp( 0)$

= $1$

So, since  $\lim_{x \to 0^{+}} x^{x}$ = 1, that means that $0^{0}$ = 1.

High School Teacher:

Showing that $x^{x}$ approaches 1 as the positive value x gets arbitrarily close to zero does not prove that $0^{0} = 1$. The variable x having a value close to zero is different than it having a value of exactly zero. It turns out that $0^{0}$ is undefined. $0^{0}$ does not have a value.

Calculus Teacher:

For all $x>0$, we have

$0^{x} = 0$.

Hence,

$\lim_{x \to 0^{+}} 0^{x} = 0$

That is, as x gets arbitrarily close to $0$ (but remains positive), $0^{x}$ stays at $0$.

On the other hand, for real numbers y such that $y \ne 0$, we have that

$y^{0} = 1$.

Hence,

$\lim_{y \to 0} y^{0} = 1$

That is, as y gets arbitrarily close to $0$, $y^{0}$ stays at $1$.

Therefore, we see that the function $f(x,y) = y^{x}$ has a discontinuity at the point $(x,y) = (0,0)$. In particular, when we approach (0,0) along the line with x=0 we get

$\lim_{y \to 0} f(0,y) = 1$

but when we approach (0,0) along the line segment with y=0 and x>0 we get

$\lim_{x \to 0^{+}} f(x,0) = 0$.

Therefore, the value of $\lim_{(x,y) \to (0,0)} y^{x}$ is going to depend on the direction that we take the limit. This means that there is no way to define $0^{0}$ that will make the function $y^{x}$ continuous at the point $(x,y) = (0,0)$.

Mathematician: Zero raised to the zero power is one. Why? Because mathematicians said so. No really, it’s true.

Let’s consider the problem of defining the function $f(x,y) = y^x$ for positive integers y and x. There are a number of definitions that all give identical results. For example, one idea is to use for our definition:

$y^x$ := $1 \times y \times y \cdots \times y$

where the y is repeated x times. In that case, when x is one, the y is repeated just one time, so we get

$y^{x}$ = $1 \times y$.

However, this definition extends quite naturally from the positive integers to the non-negative integers, so that when x is zero, y is repeated zero times, giving

$y^{0}$ = $1$

which holds for any y. Hence, when y is zero, we have

$0^0 = 1$.

Look, we’ve just proved that $0^0 = 1$! But this is only for one possible definition of $y^x$. What if we used another definition? For example, suppose that we decide to define $y^x$ as

$y^x$ := $\lim_{z \to x^{+}} y^{z}$.

In words, that means that the value of $y^x$ is whatever $y^z$ approaches as the real number z gets smaller and smaller approaching the value x arbitrarily closely.

[Clarification: a reader asked how it is possible that we can use $y^z$ in our definition of $y^x$, which seems to be recursive. The reason it is okay is because we are working here only with $z>0$, and everyone agrees about what $y^z$ equals in this case. Essentially, we are using the known cases to construct a function that has a value for the more difficult x=0 and y=0 case.]

Interestingly, using this definition, we would have

$0^0$ = $\lim_{x \to 0^{+}} 0^{x}$ = $\lim_{x \to 0^{+}} 0$ = $0$

Hence, we would find that $0^0 = 0$ rather than $0^0 = 1$. Granted, this definition we've just used feels rather unnatural, but it does agree with the common sense notion of what $y^x$ means for all positive real numbers x and y, and it does preserve continuity of the function as we approach x=0 and y=0 along a certain line.

So which of these two definitions (if either of them) is right? What is $0^0$ really? Well, for x>0 and y>0 we know what we mean by $y^x$. But when x=0 and y=0, the formula doesn't have an obvious meaning. The value of $y^x$ is going to depend on our preferred choice of definition for what we mean by that statement, and our intuition about what $y^x$ means for positive values is not enough to conclude what it means for zero values.

But if this is the case, then how can mathematicians claim that $0^0=1$? Well, merely because it is useful to do so. Some very important formulas become less elegant to write down if we instead use $0^0=0$ or if we say that $0^0$ is undefined. For example, consider the binomial theorem, which says that:

$(a+b)^x$ = $\sum_{k=0}^{\infty} \binom{x}{k} a^k b^{x-k}$

where $\binom{x}{k}$ means the binomial coefficients.

Now, setting a=0 on both sides and assuming $b \ne 0$ we get

$b^x$

= $(0+b)^x$ = $\sum_{k=0}^{\infty} \binom{x}{k} 0^k b^{x-k}$

= $\binom{x}{0} 0^0 b^{x} + \binom{x}{1} 0^1 b^{x-1} + \binom{x}{2} 0^2 b^{x-2} + \hdots$

= $\binom{x}{0} 0^0 b^{x}$

= $0^0 b^{x}$

where, I've used that $0^k = 0$ for k>0, and that  $\binom{x}{0} = 1$. Now, it so happens that the right hand side has the magical factor $0^0$. Hence, if we do not use $0^0 = 1$ then the binomial theorem (as written) does not hold when a=0 because then $b^x$ does not equal $0^0 b^{x}$.

If mathematicians were to use $0^0 = 0$, or to say that $0^0$ is undefined, then the binomial theorem would continue to hold (in some form), though not as written above. In that case though the theorem would be more complicated because it would have to handle the special case of the term corresponding to k=0. We gain elegance and simplicity by using $0^0 = 1$.

There are some further reasons why using $0^0 = 1$ is preferable, but they boil down to that choice being more useful than the alternative choices, leading to simpler theorems, or feeling more "natural" to mathematicians. The choice is not "right", it is merely nice.