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Solving $3^n fracpi^n/2(n/2)!(1+e^-pi) 2^-frac3n2 le 2^-n/2+8$
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Is it possible to solve the following inequality explicitly? Or at least proving it with some method?
$$3^n fracpi^n/2(n/2)!(1+e^-pi) 2^-frac3n2 le 2^-n/2+8$$
For $n in mathbbN$.
I am reading a research paper where they leave this up to the reader and I couldn't find a way to separate the variables, or even with Stirling's approximation. Also taking the derivative of the function results in a mess.
Graphing the functions makes the inequality seem plausible (the dots are $f(a)$, and the dashes are $f(b))$:
calculus algebra-precalculus inequality
asked Jul 31 at 0:22
Maximus
13719
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up vote
4
down vote
favorite
Is it possible to solve the following inequality explicitly? Or at least proving it with some method?
$$3^n fracpi^n/2(n/2)!(1+e^-pi) 2^-frac3n2 le 2^-n/2+8$$
For $n in mathbbN$.
I am reading a research paper where they leave this up to the reader and I couldn't find a way to separate the variables, or even with Stirling's approximation. Also taking the derivative of the function results in a mess.
Graphing the functions makes the inequality seem plausible (the dots are $f(a)$, and the dashes are $f(b))$:
calculus algebra-precalculus inequality
asked Jul 31 at 0:22
Maximus
13719
How about showing that $lim_nrightarrowinfty fracf(n)g(n)=0$
– phandaman
Jul 31 at 0:45
This has to be true for all $n in mathbbN$. I'm not exactly sure how the end behavior would impact the solution?
– Maximus
Jul 31 at 0:48
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Is it possible to solve the following inequality explicitly? Or at least proving it with some method?
$$3^n fracpi^n/2(n/2)!(1+e^-pi) 2^-frac3n2 le 2^-n/2+8$$
For $n in mathbbN$.
I am reading a research paper where they leave this up to the reader and I couldn't find a way to separate the variables, or even with Stirling's approximation. Also taking the derivative of the function results in a mess.
Graphing the functions makes the inequality seem plausible (the dots are $f(a)$, and the dashes are $f(b))$:
calculus algebra-precalculus inequality
asked Jul 31 at 0:22
Maximus
13719
Is it possible to solve the following inequality explicitly? Or at least proving it with some method?
$$3^n fracpi^n/2(n/2)!(1+e^-pi) 2^-frac3n2 le 2^-n/2+8$$
For $n in mathbbN$.
I am reading a research paper where they leave this up to the reader and I couldn't find a way to separate the variables, or even with Stirling's approximation. Also taking the derivative of the function results in a mess.
Graphing the functions makes the inequality seem plausible (the dots are $f(a)$, and the dashes are $f(b))$:
calculus algebra-precalculus inequality
asked Jul 31 at 0:22
Maximus
13719
edited Jul 31 at 0:27
asked Jul 31 at 0:22
Maximus
13719
asked Jul 31 at 0:22
Maximus
13719
asked Jul 31 at 0:22
Maximus
13719
13719
How about showing that $lim_nrightarrowinfty fracf(n)g(n)=0$
– phandaman
Jul 31 at 0:45
This has to be true for all $n in mathbbN$. I'm not exactly sure how the end behavior would impact the solution?
– Maximus
Jul 31 at 0:48
add a comment |Â
How about showing that $lim_nrightarrowinfty fracf(n)g(n)=0$
– phandaman
Jul 31 at 0:45
This has to be true for all $n in mathbbN$. I'm not exactly sure how the end behavior would impact the solution?
– Maximus
Jul 31 at 0:48
How about showing that $lim_nrightarrowinfty fracf(n)g(n)=0$
– phandaman
Jul 31 at 0:45
How about showing that $lim_nrightarrowinfty fracf(n)g(n)=0$
– phandaman
Jul 31 at 0:45
This has to be true for all $n in mathbbN$. I'm not exactly sure how the end behavior would impact the solution?
– Maximus
Jul 31 at 0:48
This has to be true for all $n in mathbbN$. I'm not exactly sure how the end behavior would impact the solution?
– Maximus
Jul 31 at 0:48
add a comment |Â
1 Answer
1
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0
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Hint: Your inequality is equivalent to: $$dfracpi^n/2(n/2)!(1+e^-pi)leq 256left(frac23right)^n$$
or $$f(n) = dfrac(2.25cdotpi)^n/2(n/2)!leqdfrac2561+e^-pi=245.395...$$
Now, you can simply observe that the LHS is eventually decreasing eventually because a factorial is increasing more rapidly than an exponent. This means there is a single, global maximum at some $n = k$, where it has to satisfy the inequalities: $$f(k-1)<f(k)$$ and $$f(k)>f(k+1).$$
If you solve these inequalities, then you can easily find that $k = 14$, which gives the maximum of $f(k) = 174.943.$
answered Jul 31 at 4:09
dezdichado
5,4591826
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Hint: Your inequality is equivalent to: $$dfracpi^n/2(n/2)!(1+e^-pi)leq 256left(frac23right)^n$$
or $$f(n) = dfrac(2.25cdotpi)^n/2(n/2)!leqdfrac2561+e^-pi=245.395...$$
Now, you can simply observe that the LHS is eventually decreasing eventually because a factorial is increasing more rapidly than an exponent. This means there is a single, global maximum at some $n = k$, where it has to satisfy the inequalities: $$f(k-1)<f(k)$$ and $$f(k)>f(k+1).$$
If you solve these inequalities, then you can easily find that $k = 14$, which gives the maximum of $f(k) = 174.943.$
answered Jul 31 at 4:09
dezdichado
5,4591826
add a comment |Â
up vote
0
down vote
accepted
Hint: Your inequality is equivalent to: $$dfracpi^n/2(n/2)!(1+e^-pi)leq 256left(frac23right)^n$$
or $$f(n) = dfrac(2.25cdotpi)^n/2(n/2)!leqdfrac2561+e^-pi=245.395...$$
Now, you can simply observe that the LHS is eventually decreasing eventually because a factorial is increasing more rapidly than an exponent. This means there is a single, global maximum at some $n = k$, where it has to satisfy the inequalities: $$f(k-1)<f(k)$$ and $$f(k)>f(k+1).$$
If you solve these inequalities, then you can easily find that $k = 14$, which gives the maximum of $f(k) = 174.943.$
answered Jul 31 at 4:09
dezdichado
5,4591826
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Hint: Your inequality is equivalent to: $$dfracpi^n/2(n/2)!(1+e^-pi)leq 256left(frac23right)^n$$
or $$f(n) = dfrac(2.25cdotpi)^n/2(n/2)!leqdfrac2561+e^-pi=245.395...$$
Now, you can simply observe that the LHS is eventually decreasing eventually because a factorial is increasing more rapidly than an exponent. This means there is a single, global maximum at some $n = k$, where it has to satisfy the inequalities: $$f(k-1)<f(k)$$ and $$f(k)>f(k+1).$$
If you solve these inequalities, then you can easily find that $k = 14$, which gives the maximum of $f(k) = 174.943.$
answered Jul 31 at 4:09
dezdichado
5,4591826
Hint: Your inequality is equivalent to: $$dfracpi^n/2(n/2)!(1+e^-pi)leq 256left(frac23right)^n$$
or $$f(n) = dfrac(2.25cdotpi)^n/2(n/2)!leqdfrac2561+e^-pi=245.395...$$
Now, you can simply observe that the LHS is eventually decreasing eventually because a factorial is increasing more rapidly than an exponent. This means there is a single, global maximum at some $n = k$, where it has to satisfy the inequalities: $$f(k-1)<f(k)$$ and $$f(k)>f(k+1).$$
If you solve these inequalities, then you can easily find that $k = 14$, which gives the maximum of $f(k) = 174.943.$
answered Jul 31 at 4:09
dezdichado
5,4591826
answered Jul 31 at 4:09
dezdichado
5,4591826
answered Jul 31 at 4:09
dezdichado
5,4591826
answered Jul 31 at 4:09
dezdichado
5,4591826
5,4591826
add a comment |Â
add a comment |Â
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How about showing that $lim_nrightarrowinfty fracf(n)g(n)=0$
– phandaman
Jul 31 at 0:45
This has to be true for all $n in mathbbN$. I'm not exactly sure how the end behavior would impact the solution?
– Maximus
Jul 31 at 0:48