## 400 Reputation

16 years, 325 days

## solution may have been lost...

`> 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`

## solution may have been lost...

`> 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. ...

hi hirnyk 1271 thanks.

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

and

we get

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

## hi hirnyk 1271 thanks. ...

hi hirnyk 1271 thanks.

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

and

we get

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

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

1

m:= 1

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

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

1

m:= 1

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 const...

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 const...

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 ri...

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 ri...

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 ...

thank you joe. it looks like an interesting insight.

## thank you joe. it looks like an ...

thank you joe. it looks like an interesting insight.

## absolutely sir. have i told...

absolutely sir.

have i told u: u make things look beautiful.

## absolutely sir. have i told...

absolutely sir.

have i told u: u make things look beautiful.

## Thanks Dough. You are...

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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