Lonely

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15 years, 13 days

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These are replies submitted by Lonely

 

i think the series would converge:

 

sum((1+1/n)^n - exp(1),n=1..infinity)

sum((1+1/n)^(n+0.5) -exp(1),n=1..infinity)

 

let us consider the question: for what value of a, (1+1/n)^(n+a) is nearest to exp(1).

The answer is a = 1/2 ( i think)

Now, as suggested by you. I tried:

> q := proc (n, a) options operator, arrow; ((1+1/(n+1))^(n+1+a)-exp(1))/((1+1/n)^(n+a)-exp(1)) end proc

> assume(a,'real')

> taylor(q(n, a), n = infinity, 5)

gives the following:-

1+(exp(1)*(23/24-2*a+(1/2)*a^2)-exp(1)*(11/24-a+(1/2)*a^2))/(exp(1)*(a-1/2)*n)+(exp(1)*(-89/48+(97/24)*a-(7/4)*a^2+(1/6)*a^3)-exp(1)*(-7/16+(25/24)*a-(3/4)*a^2+(1/6)*a^3)+exp(1)*(11/24-a+(1/2)*a^2))/(exp(1)*(a-1/2)*n^2)+O(1/n^3)

Now we can see (by hand calculation, i donot know how to simplifiy it by maple) that the second term of the above series is: -1/n

which is independent of a.

While if we find out the taylor series for a = 0.5, we see that second term is -2/n . I mean if we find the series of :

taylor(q(n, 1/2), n = infinity, 6)

we get

1-2/n+4/n^2+O(1/n^3)

 

 

let us consider the question: for what value of a, (1+1/n)^(n+a) is nearest to exp(1).

The answer is a = 1/2 ( i think)

Now, as suggested by you. I tried:

> q := proc (n, a) options operator, arrow; ((1+1/(n+1))^(n+1+a)-exp(1))/((1+1/n)^(n+a)-exp(1)) end proc

> assume(a,'real')

> taylor(q(n, a), n = infinity, 5)

gives the following:-

1+(exp(1)*(23/24-2*a+(1/2)*a^2)-exp(1)*(11/24-a+(1/2)*a^2))/(exp(1)*(a-1/2)*n)+(exp(1)*(-89/48+(97/24)*a-(7/4)*a^2+(1/6)*a^3)-exp(1)*(-7/16+(25/24)*a-(3/4)*a^2+(1/6)*a^3)+exp(1)*(11/24-a+(1/2)*a^2))/(exp(1)*(a-1/2)*n^2)+O(1/n^3)

Now we can see (by hand calculation, i donot know how to simplifiy it by maple) that the second term of the above series is: -1/n

which is independent of a.

While if we find out the taylor series for a = 0.5, we see that second term is -2/n . I mean if we find the series of :

taylor(q(n, 1/2), n = infinity, 6)

we get

1-2/n+4/n^2+O(1/n^3)

 

i very much appreciate your help. in your direction, i tried the following:


Q1 := n->(1+1/n)^(n+a) assuming a real:

R2 :=(Q1(n+1)-exp(1))/(Q1(n)-exp(1)):

 

taylor(R2,n=infinity,5)

 

the second term of the above series does not contains a.  Even if I substitute a = 0.5, I donot get back the series you generated through

series( r2, n=infinity );

i very much appreciate your help. in your direction, i tried the following:


Q1 := n->(1+1/n)^(n+a) assuming a real:

R2 :=(Q1(n+1)-exp(1))/(Q1(n)-exp(1)):

 

taylor(R2,n=infinity,5)

 

the second term of the above series does not contains a.  Even if I substitute a = 0.5, I donot get back the series you generated through

series( r2, n=infinity );

thanks.

if we give the command:

g(x) := arctan( cos(2*x) ) + Pi/4;

int(g(x), x = 0 .. 2);

maple gives answer which is very complex to me can we simplify that

 

thanks.

if we give the command:

g(x) := arctan( cos(2*x) ) + Pi/4;

int(g(x), x = 0 .. 2);

maple gives answer which is very complex to me can we simplify that

 

we may assume 0<a<b

we may assume 0<a<b

thanks.

what is SCR?

thanks.

what is SCR?

 

int(exp(-xi^2*(3*n+xi)/(6*n^2+6*xi*n+xi^2)), xi = -infinity .. infinity)

We may assume (xi/n ->0)

 

 

int(exp(-xi^2*(3*n+xi)/(6*n^2+6*xi*n+xi^2)), xi = -infinity .. infinity)

We may assume (xi/n ->0)

 

i think the integral exists.

can we find a closed form solution to this integral?

i think the integral exists.

can we find a closed form solution to this integral?

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