Items tagged with exponential exponential Tagged Items Feed

Hi,

My question is fairly straight-forward, I hope someone has an equally straight-forward solution. I would like to be able to illustrate the convergence of the distribution on the mean of i.i.d random variables to the normal distribution. For example, I'd like to visualize (by way of a histogram plot) that as n increases arbitrarily, the mean of n i.i.d. exponential random variables with mean 10 converges to a normal distribution.

Thanks!

I have a solution containing many exponentail terms, some of which are in the denominators of rational terms. I would like to be able to have the solution given to me in a manner where there are no exponentials in denominators but only in the numerators. The simplify command in maple does it, however for the shear number of terms (just shy of 400,000 terms and maple saying it's million plus) i am looking at, that particular command is just taking too long (15 hours!). Is there...

How to make maple to change ln(exp(x)) = x?  for example exp(ln(x)) = x, or ln(exp(x+1)) = x+1

One thing I have always found interesting is that exponential growth/decay
is not symmetrical hence:

100*(1+0.01)^2+100*(1-0.01)^2-200 = 0.0200

For example:

If you start with 100 and bet 1% and you win and then again bet 1% of your portfolio value
and you win again your portfolio value will be PW=100*(1+0.01)^2.

If you start with 100 and bet 1% and you lose and then again bet 1% of your portfolio value
and you lose again your portfolio value will be PL=100*(1-0.01)^2. ...

Hi,

I am trying to simplify the expression s as given below. (I am not sure why it comes up with all the vector caclulus notation in it but it should display okay when you enter it)

Because of the presence of the exponential imaginary fucntions I thought evalc might be useful but when I use it I get a huge expression with csgn appearing in it. To my knowledge csgn appears when assumptions are not correctly specified - is this so? I can't see any assumption...

Why doesn't 'ln(e)' simplify to '1' in Maple 12.02?

The problem arose from trying to simplify the solution of an ODE that contains a combination of exponentials, logarithms and powers. Maple refuses to simplify, whereas I can easily read the simplified answer. Maple seems to ignore the simplest of algebra concerning power, logarithms and exponentials.

Hi

I'm trying to find the argument (or phase) of the following expression:

where r,d, and t are all positive real numbers. w is also a real number (although could be negative).

Can anyone give me hints as to the best way to do this in Maple? Thanks for thinking...

How to interpolate in terms of any form of algebra in maple?

For example, to interpolate a time series in terms of exponential or mixed with cos and sin?

Included below is vector partial diff eq I am working with.  To get rid of the time deriv's I took the LaPlace transform & the remaining spatial eq in the s-domain is listed.  To make matters simpler I set beta = 0 to get rid of the curl of the field.  What remains is essentially the Helmholtz eq.  To simplify further I just found the homogeneous soln for the x direction only.

As can be seen the eigenfunctions are exponentials with s beneath...

What it is wrong with this one-line-document:

with(Optimization): nx := Maximize(e^x, {x <= 1.})

that give this answer:

[9.99990000000000000*10^19, [e = 9.99990000000000000*10^19, x = 1.]]

I have a complicated function (combination of exponential and sin), and I want to find the infinite integral of that from 0 to infinity which I know that the function itself converges to zero at infinity. I used evalf to force maple to do that but it fails to give the numerical answer. How can I resolve the problem?

Thanks,

Hamid

I need help . Why won't Maple evaluate the 2nd limit? How do I get around that? T is a constant.

> limit(exp(x/2),x=infinity);

                               infinity


> limit(exp(x/T),x=infinity);

                ...

How do I solve double exponential problem of the form  : 

 i = k_c* i_p *( e ^-a₁*t - e ^-a₂*t ) given that for d/dt (i) =0. t_p = ( i/(a₂ -a₁)) *ln(a₂/a₁) and                                

k_c = 1/(e ^{-a₁ *t_p}

Hi,

I am trying to force the conversion of

to its cleaner looking

Anybody know how to do this in a straight-forward way? I've tried convert...

Hello everyone!

I am stuck after browsing my book and the help section for some time.

How can I find the value of a term like (with MAPLE syntax)

for x=1,2,..,5 ?

I'm looking for something like matlab's k=1:5; operation.

Creating sequences or arrays/vectors didn't help.

Thank you for your time if you read this.