mschneider

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2 years, 175 days

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These are questions asked by mschneider

Please help...

answer := 5*exp(7.5*t);

response := subs(a = exp(1), 5*a^(7.5*t));

a := evalf(subs(t = Pi, answer));

b := evalf(subs(t = Pi, response));

evalf(a-b);

 

Any explanations as to why the last line is not zero?  any workarounds?

I would like to perform an implicit differentiation of the following:

eqn := 5*(x*y)^2+x/sqrt(y) = x^2+2*(3*x^3+y^2)^3

However implicitdiff(eqn, x, y); produces an output that expands the result.  Is there a way to not have it expand?

I tried something to the effect of...

eqn := 5*(x*y(x))^2+x/sqrt(y(x)) = x^2+2*(3*x^3+y(x)^2)^3;

diff(eqn, x);

but now what I need to to isolate d/dx y(x)...

Any suggestions are much appreciated.

I am attempting to have maple recognize the difference in singularities between:

f(x)=x+1  and

g(x)=(x+1)*(x-2)/(x-2)

In other words how can I stop maple from simplifying g(x) and observe the singularity.

 

Thanks in advance.

Mark

 

I am looking to develop a histogram from a random data set that contains integers that are approximately normally distributed.  The intent is to develop a histogram from the dataset that looks like a fairly decent-approximation of a normalized curve (but not necessarily perfect)....  the next stage to this is a second histogram that shows a bimodal distribution (but not necessarily perfect)

 

To this point I have...

restart;

with(Statistics);

randomize();

N := RandomVariable(Normal(300, 10));

A := [seq(MapleTA:-Builtin:-decimal(0, Sample(N, 500)[i]), i = 1 .. 500)];

Histogram(A, frequencyscale = absolute, bincount = 25, binwidth = 1, tickmarks = [default, default]);

 

I am certain you will find what I have thus far is far from ideal, as such any help is appreciated.

 

Thanks,

Mark

 

Hello,

Can someone explain to me why the following lists do not have the same content?  And suggest how I might be able to fix the problem?

 

restart;

a := (x^2+y^2)^3 = y^2:

b := (x^2+y^2)^(3/2) = y:

listA := sort([seq(evalf(solve({y >= 0, subs(x = i, a)}, {y})), i = 0.5e-1 .. .5, 0.5e-1)]);

listB := sort([seq(evalf(solve({y >= 0, subs(x = i, b)}, {y})), i = 0.5e-1 .. .5, 0.5e-1)]);

evalb(listA = listB);

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