Thanks Dough.

You are absolutely correct.

How to do it with Maple?

Happy new year and wishing all the happiness in coming years

thanks for your help.

how do you get this answer?

Are there functions ? Certainly there are ?

thanks for your help.

how do you get this answer?

Are there functions ? Certainly there are ?

hi robert thanks for your help.

how can be expand the following:

1. suppose we have a taylor series of function f(x)

A := taylor(f(x), x = gamma, n)

2. How can we find a series of

A^m

we may assume m::posint

hi robert thanks for your help.

how can be expand the following:

1. suppose we have a taylor series of function f(x)

A := taylor(f(x), x = gamma, n)

2. How can we find a series of

A^m

we may assume m::posint

thanks

i need when the boundary conditions are given as a function in time:

for example T(0,t) = f(t) and T(l,t) = p(t) and initial temperature distribution may be assumed constant

thanks

i need when the boundary conditions are given as a function in time:

for example T(0,t) = f(t) and T(l,t) = p(t) and initial temperature distribution may be assumed constant

Thanks Robert.

Would you please explain further how to test the convergence of the given series with Maple?

Thanks Robert.

Would you please explain further how to test the convergence of the given series with Maple?

it can be seen graphicall that

p in [0,3] inequality is satisfied

but how to show it

it can be seen graphicall that

p in [0,3] inequality is satisfied

but how to show it

(1+x)^(p+1) > 2^p*(x^p+x)

here 1 < = x < = 2

(1+x)^(p+1) > 2^p*(x^p+x)

here 1 < = x < = 2

just a wild idea: if we see:- the first part of the series:

int(1/(x*(1+x^1/3),x=1..infinity)

int(((x+1)/x-1)/(1+x^(1/3)), x = 1 .. infinity)

= 3* ln(2)

just a wild idea: if we see:- the first part of the series:

int(1/(x*(1+x^1/3),x=1..infinity)

int(((x+1)/x-1)/(1+x^(1/3)), x = 1 .. infinity)

= 3* ln(2)