## 185 Reputation

10 years, 64 days

## On listlists...

Thanks for all the replies!  How do I carry out those operations on listlists?

L:=[[1,2,3],[4,3,2]]

map(`^`,L,2)

I prefer [[1,4,9],[16,9,4]] as the answer.

## Thanks...

@DSkoog Yes thanks that helps, it works in 2015 but does not in Maple 18 (was that a bug for 18?)

## Thanks...

Thanks for all the inputs.

I may not be thinking about the problem in the most efficient manner but all three answers above should work for me.
Each pairing represents a team and a score.

## In mathematica but how about Maple...

In mathematica to turn any data time series into sound you can do this.

``data =WeatherData["Boston","MeanTemperature",{{1970},{2012},"Day"}];``
``DateListLogPlot[data,PlotStyle->PointSize[0],AspectRatio->1/5]``

``ListPlay[data[[All,2]],SampleRate->2000]``

To do in Maple I'm not sure.

## odds of dying...

Thanks for the inputs.

So lets say the branch break automatically results in a fall.  From the 2015 injury facts http://www.nsc.org/learn/safety-knowledge/Pages/injury-facts-chart.aspx the odds of someone dying in a fall are 1:144

So in the simulation we would just replace the probability with P(branch break)*P(dying in a fall) (.02)*(1/144) to find the probablity of someone dying in such a situation.

We could break that down further and find the probability of death falling from a certain height, so as someone climbs higher and higher in that tree the situation becomes increasingly more grave.  I couldn't find any statistics related to death vs. height but I did find that a height above 5 stories, death is almost 100% certain.

## nice...

@acer I love the 3d Maple.  Thanks Acer!

I see using text in your first example it needed to be converted into a graphic first.

## Thanks...

@Carl Love thanks.  That was simple enough, I way overlooked that one.  Thanks again.

## Thanks...

@Markiyan Hirnyk Thank you

## area under curve...

Is the area under the curve 100% in your first graph and the last one that was produced?

Or because it is skewed and truncated it is not ?

## Maybe Poisson...

That's great!  Regarding your first plot, that seems to be the closest to what I'm after.

Is the probability under the curve almost 100%?

This Poisson distribution also seems close

P(x)= (exp(-lambda)*lambda^x) / x!    # for lambda=2

## particulars aside...

@Markiyan Hirnyk ok well it may not be a truncated normal.  But set aside all that for a moment.  I may not be versed in the statistics lingo enough to understand what it's called.

Let's take a normal distribution chop the left side of it off around 3/4 the way up and shift it over to the y-axis.  Then stretch the other side of the normal over to 15 while keeping the mode (peak) near or at 1 doesn't really matter.  Maybe we could call it a truncated skewed normal at least in laymans terms that's what I would call it.

How can we set up calculations to show a graph like that.

## A picture...

A picture like this

Where 1 min has the highest of probabilities (40%), immediately served (time 0) and trailing off to 15minutes after 1 minute.  However I don't think an exponential distribution would fit.

## Patch library...

Would this not also qualify as a patch library?

## Physics library...

I suppose even the ever updating Physics library is a similar idea to this Maple patch library.  And a very successful one at that.

## euler lagrange...

Thanks, so the ode is essentially the euler-lagrange equation?

After using    L := -sin(omega*t-phi(t))*(diff(phi(t), t))*a*l*m*omega+(1/2)*(diff(phi(t), t))^2*l^2*m+cos(phi(t))*g*l*m

I tried to diff(L,diff(phi(t)))  for the second part in the euler lagrange equation and it threw an error, Error, invalid input: diff received diff(phi(t), t), which is not valid for its 2nd argument

I am not sure why.  Maybe L is supposed to be a function of phi ?

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