From: Flora giannone on 26 Nov 2009 07:36 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
From: Thomas Arildsen on 26 Nov 2009 07:56 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 -- All email to sender address is lost. My real adress is at es dot aau dot dk for user tha.
From: Flora giannone on 26 Nov 2009 08:55 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?
From: Thomas Arildsen on 26 Nov 2009 09:17 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.
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