Thread Subject: help,Symbolic math toolbox

hi guys
i have a problem with the symbolic math toolbox.

I have a function
a=inline('(n+1)/n');
I want to prove that a (n)> a (n +1)
but this can not be calculated with the symbolic math toolbox.
how can I do?????
thanks
Flo

On Thu, 26 Nov 2009 12:36:03 +0000, Flora giannone wrote:

> hi guys
> i have a problem with the symbolic math toolbox.
>
> I have a function
> a=inline('(n+1)/n');
> I want to prove that a (n)> a (n +1)
> but this can not be calculated with the symbolic math toolbox. how can I
> do?????
> thanks
> Flo

How about:
(n+1)/n > (n+2)/(n+1)
(n+1)^2/n > (n+2)
(n+1)^2 > n*(n+2)
n^2 + 2*n + 1 > n^2 + 2*n
1 > 0
q.e.d.
In the above steps, I assumed n>0 for simplicity. I don't know which
conditions you have on n.

Thomas Arildsen

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thanks thomas for your answer...
the problem is that i have more functions inline;

M=menu('Scegliere una serie','z^k','z^k/k','z^k/k!',...
'z^n/2^2n','(z+1)^n/(n*2^n)','(z-2)^n/(n^2*2^2n)','(1+ni)z^n',...
'log(n)^2*z^n','5^n*z^3n/2n*(2n+2)');
%Definisco i coefficienti
switch M
    case 1 %serie Geometrica
        a=inline(sym(1));
        %La serie geometrica ha coefficiente = 1
        x=0;
        %Ci serve per calcolare la somma dei termini della serie
        an=z^n;
        %Termine generale della serie
    case 2 %Serie Armonica
        a=inline('(n+1)/n');
        x=1;
        an=(z^n)/n;
    case 3 %Serie Esponenziale
      a=inline('1/gamma(n+1)');
        x=0;
        an=(z^n)/gamma(n+1);
    case 4
        a=inline('1/4^n');
        x=0;
        an=(z^n)/(4^n);

I do not want to do these calculations manually...there is no function that allows me to avoid these calculations?

On Thu, 26 Nov 2009 13:55:07 +0000, Flora giannone wrote:

> thanks thomas for your answer...
> the problem is that i have more functions inline;
>
> M=menu('Scegliere una serie','z^k','z^k/k','z^k/k!',...
> 'z^n/2^2n','(z+1)^n/(n*2^n)','(z-2)^n/(n^2*2^2n)','(1+ni)z^n',...
> 'log(n)^2*z^n','5^n*z^3n/2n*(2n+2)'); %Definisco i coefficienti
> switch M
> case 1 %serie Geometrica
> a=inline(sym(1));
> %La serie geometrica ha coefficiente = 1 x=0;
> %Ci serve per calcolare la somma dei termini della serie an=z^n;
> %Termine generale della serie
> case 2 %Serie Armonica
> a=inline('(n+1)/n');
> x=1;
> an=(z^n)/n;
> case 3 %Serie Esponenziale
> a=inline('1/gamma(n+1)');
> x=0;
> an=(z^n)/gamma(n+1);
> case 4
> a=inline('1/4^n');
> x=0;
> an=(z^n)/(4^n);
>
> I do not want to do these calculations manually...there is no function
> that allows me to avoid these calculations?

That I don't know. I only considered your first quite simple example,
sorry.

Thomas Arildsen

--
All email to sender address is lost.
My real adress is at es dot aau dot dk for user tha.