Lonely

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15 years, 78 days

MaplePrimes Activity


These are replies submitted by Lonely

 

> restart;
> eq1 := (4-(3/2)*q+(1/2)*sqrt(28-24*q+5*q^2))*(1/2-(1/2)*q);
> eq2 := (3-q+(1/4)*sqrt(84-58*q+10*q^2)+(1/4)*sqrt((4-2*q)*(3-q)))*(1/2-(1/2)*q);
> solve({eq1 < 1, eq2 < 1}, {q});
Warning, solutions may have been los

 

> restart;
> eq1 := (4-(3/2)*q+(1/2)*sqrt(28-24*q+5*q^2))*(1/2-(1/2)*q);
> eq2 := (3-q+(1/4)*sqrt(84-58*q+10*q^2)+(1/4)*sqrt((4-2*q)*(3-q)))*(1/2-(1/2)*q);
> solve({eq1 < 1, eq2 < 1}, {q});
Warning, solutions may have been los

 

 

hi hirnyk 1271 thanks.

 

let us do as suggested by helge at mo. define a function g(n,m) = ln(f(n,m))

 

g(n,m):=1/(2)*(sum(1/((2*k+1)*(2*n+1)^(2*k+1)), k = 0 .. m))^(-1)*ln(1+1/(n))

 

and

 

asympt(g(n, m), n)

 

we get

 

1+O(1/n^3)

 

what does it tell us? How can I see the dependence of O(?) on m

 

 

 

hi hirnyk 1271 thanks.

 

let us do as suggested by helge at mo. define a function g(n,m) = ln(f(n,m))

 

g(n,m):=1/(2)*(sum(1/((2*k+1)*(2*n+1)^(2*k+1)), k = 0 .. m))^(-1)*ln(1+1/(n))

 

and

 

asympt(g(n, m), n)

 

we get

 

1+O(1/n^3)

 

what does it tell us? How can I see the dependence of O(?) on m

 

 

 

thank you so much to all of you. as pointed out by robert

sum(1/((2*k+1)*(2*n+1)^(2*k+1)), k = 0 .. infinity)) = 1/2 ln(1+1/n)

Now

T := exp(1)-(1+1/n)^((1/2)/(sum(1/((2*k+1)*(2*n+1)^(2*k+1)), k = 0 .. m)))

now to find out BigO(1/n^f(m))  if we do the following

1

m:= 1

series(T, n = infinity)*assume*m > 0

we will get O(1/n^4)

2.

m:=2 

series(T, n=infinity,5)

we will get 0(1/n^6)

3. for m= 3 we will get O(n^8)

4. m = 4 we will get O(1/n^10)

 

thus it looks like it is behaving like O(1/n^(2*m+2))

 

now how can we show that for T the asymptotic rate is O(1/n^(2*m+2))

 

 

 

thank you so much to all of you. as pointed out by robert

sum(1/((2*k+1)*(2*n+1)^(2*k+1)), k = 0 .. infinity)) = 1/2 ln(1+1/n)

Now

T := exp(1)-(1+1/n)^((1/2)/(sum(1/((2*k+1)*(2*n+1)^(2*k+1)), k = 0 .. m)))

now to find out BigO(1/n^f(m))  if we do the following

1

m:= 1

series(T, n = infinity)*assume*m > 0

we will get O(1/n^4)

2.

m:=2 

series(T, n=infinity,5)

we will get 0(1/n^6)

3. for m= 3 we will get O(n^8)

4. m = 4 we will get O(1/n^10)

 

thus it looks like it is behaving like O(1/n^(2*m+2))

 

now how can we show that for T the asymptotic rate is O(1/n^(2*m+2))

 

 

 

why the asymptotic constant is independet of m

 

the asymptotic is O(1/n^(2m+2))

 

but how to find it with maple

 

 

why the asymptotic constant is independet of m

 

the asymptotic is O(1/n^(2m+2))

 

but how to find it with maple

 

 

thank you and you are right it ain't looking pretty.

 

any idea to make it look nice (manageable) by some approximation

 

 

thank you and you are right it ain't looking pretty.

 

any idea to make it look nice (manageable) by some approximation

 

thank you joe. it looks like an interesting insight.

 

thank you joe. it looks like an interesting insight.

absolutely sir.

have i told u: u make things look beautiful.

absolutely sir.

have i told u: u make things look beautiful.

Thanks Dough.

You are absolutely correct.

How to do it with Maple?

Happy new year and wishing all the happiness in coming years

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