Lonely

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These are questions asked by Lonely

<p> </p>
<p><img alt="" src="file:///C:/DOCUME~1/sanjay/LOKALE~1/Temp/moz-screenshot-18.jpg" /><img alt="" src="file:///C:/DOCUME~1/sanjay/LOKALE~1/Temp/moz-screenshot-19.jpg" /><maple>e < proc (n) options operator, arrow; (1+1/n)^((1/8)*(n^(1/3)+(n+1)^(1/3))^3) end proc</maple></p>

hi

how can i find the convergence rate of the sequence:

 

(1+1/n)^n

 

how can i show that

(1+1/n)^(n+1/2) converges faster than (1+1/n)^n

 

Hi,

we know that:

X(n+1) =Xn - f(x_n)/f^\prime(x_n)

Now I am using the following commands for showing the 2nd order convergence of the above method:

> restart;
> f(a) :=0: 
> A := taylor(f(x), x = a, 4);
>
> B := taylor((D(f))(x), x = a, 4);
> C := series(e[n]-A/(D(f))(x), x = a, 3);
> simplify(subs(x-a = e[n], %));
 

it does not cancel e_n

I know there must be a better way of doint it. Can you please tell me?

we want to find the sum:

sum((6*n+5)*(6*n+3)/((12*n*n+12*n+1)*(6*n*n+6*n+1)), n = 1 .. infinity)

is is equal to:

-15-(1/2)*sqrt(3)*Pi*tan((1/6)*Pi*sqrt(3))+(1/2)*sqrt(6)*Pi*tan((1/6)*Pi*sqrt(6))-2*Psi(1/2+(1/6)*sqrt(6))+Pi*tan((1/6)*Pi*sqrt(6))+2*Psi(1/2+(1/6)*sqrt(3))-Pi*tan((1/6)*Pi*sqrt(3))

can we get rid of  the function "\Psi"?  Can we bound this function \Psi with other known constant such as zeta function?

 

is it imaginary:

 

evalf((-1)^(1/3))

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