## 400 Reputation

15 years, 253 days

## Thanks Dough. You are...

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...

how do you get this answer?

Are there functions ? Certainly there are ?

## thanks for your help. how do...

how do you get this answer?

Are there functions ? Certainly there are ?

## hi robert thanks for your...

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...

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...

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...

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...

Thanks Robert.

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

## thanks Robert...

Thanks Robert.

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

## p > 0...

it can be seen graphicall that

p in [0,3] inequality is satisfied

but how to show it

## p > 0...

it can be seen graphicall that

p in [0,3] inequality is satisfied

but how to show it

## thanks and our inequality is...

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

here 1 < = x < = 2

## thanks and our inequality is...

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

here 1 < = x < = 2

## thanks again...

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)

## thanks again...

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)

 1 2 3 4 5 Page 2 of 5
﻿