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Polylogarithm grows slower than polynomial proof
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In the CLRS book, there's this part, where it's shown that $$lim_{ntoinfty}frac{(n^b)}{(a^n)} = 0$$. In the same chapter, it uses the aforementioned equation to prove that any polylogarithm function grows slower than any polynomial one, thus, $$lim_{ntoinfty}frac{log b^n}{ n^a}$$. It does that by substituting lgn for n and 2^a for a in the first equation. How is it allowed to substitute the terms and prove the latter equation.
polynomials polylogarithm
edited Feb 1 at 11:18
mm-crj
423213
asked Feb 1 at 10:53
chakmeshmachakmeshma
1203
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add a comment |
$begingroup$
In the CLRS book, there's this part, where it's shown that $$lim_{ntoinfty}frac{(n^b)}{(a^n)} = 0$$. In the same chapter, it uses the aforementioned equation to prove that any polylogarithm function grows slower than any polynomial one, thus, $$lim_{ntoinfty}frac{log b^n}{ n^a}$$. It does that by substituting lgn for n and 2^a for a in the first equation. How is it allowed to substitute the terms and prove the latter equation.
polynomials polylogarithm
edited Feb 1 at 11:18
mm-crj
423213
asked Feb 1 at 10:53
chakmeshmachakmeshma
1203
$endgroup$
add a comment |
$begingroup$
In the CLRS book, there's this part, where it's shown that $$lim_{ntoinfty}frac{(n^b)}{(a^n)} = 0$$. In the same chapter, it uses the aforementioned equation to prove that any polylogarithm function grows slower than any polynomial one, thus, $$lim_{ntoinfty}frac{log b^n}{ n^a}$$. It does that by substituting lgn for n and 2^a for a in the first equation. How is it allowed to substitute the terms and prove the latter equation.
polynomials polylogarithm
edited Feb 1 at 11:18
mm-crj
423213
asked Feb 1 at 10:53
chakmeshmachakmeshma
1203
$endgroup$
In the CLRS book, there's this part, where it's shown that $$lim_{ntoinfty}frac{(n^b)}{(a^n)} = 0$$. In the same chapter, it uses the aforementioned equation to prove that any polylogarithm function grows slower than any polynomial one, thus, $$lim_{ntoinfty}frac{log b^n}{ n^a}$$. It does that by substituting lgn for n and 2^a for a in the first equation. How is it allowed to substitute the terms and prove the latter equation.
polynomials polylogarithm
polynomials polylogarithm
edited Feb 1 at 11:18
mm-crj
423213
asked Feb 1 at 10:53
chakmeshmachakmeshma
1203
edited Feb 1 at 11:18
mm-crj
423213
asked Feb 1 at 10:53
chakmeshmachakmeshma
1203
edited Feb 1 at 11:18
mm-crj
423213
edited Feb 1 at 11:18
mm-crj
423213
edited Feb 1 at 11:18
mm-crj
423213
423213
asked Feb 1 at 10:53
chakmeshmachakmeshma
1203
asked Feb 1 at 10:53
chakmeshmachakmeshma
1203
asked Feb 1 at 10:53
chakmeshmachakmeshma
1203
1203
add a comment |
add a comment |
1 Answer
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Let $a_n:= frac{log b^n}{ n^a}$, then $a_n =frac{n log b}{ n^a}=log b frac{1}{n^{a-1}}.$
We can assume that $log b ne 0.$
Case 1: $a=1.$ Then $a_n to log b .$
Case 2: $a>1$. Then $a_n to 0.$
Case 3: $a<1.$ Then $a_n to infty$, if $b>1$ and $a_n to - infty$, if $0<b<1.$
answered Feb 1 at 11:29
FredFred
48.5k11849
$endgroup$
$begingroup$
My question is, how does this justify replacing the terms (lgn for n and a for 2^a)
$endgroup$
– chakmeshma
Feb 2 at 6:53
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Let $a_n:= frac{log b^n}{ n^a}$, then $a_n =frac{n log b}{ n^a}=log b frac{1}{n^{a-1}}.$
We can assume that $log b ne 0.$
Case 1: $a=1.$ Then $a_n to log b .$
Case 2: $a>1$. Then $a_n to 0.$
Case 3: $a<1.$ Then $a_n to infty$, if $b>1$ and $a_n to - infty$, if $0<b<1.$
answered Feb 1 at 11:29
FredFred
48.5k11849
$endgroup$
$begingroup$
My question is, how does this justify replacing the terms (lgn for n and a for 2^a)
$endgroup$
– chakmeshma
Feb 2 at 6:53
add a comment |
$begingroup$
Let $a_n:= frac{log b^n}{ n^a}$, then $a_n =frac{n log b}{ n^a}=log b frac{1}{n^{a-1}}.$
We can assume that $log b ne 0.$
Case 1: $a=1.$ Then $a_n to log b .$
Case 2: $a>1$. Then $a_n to 0.$
Case 3: $a<1.$ Then $a_n to infty$, if $b>1$ and $a_n to - infty$, if $0<b<1.$
answered Feb 1 at 11:29
FredFred
48.5k11849
$endgroup$
$begingroup$
My question is, how does this justify replacing the terms (lgn for n and a for 2^a)
$endgroup$
– chakmeshma
Feb 2 at 6:53
add a comment |
$begingroup$
Let $a_n:= frac{log b^n}{ n^a}$, then $a_n =frac{n log b}{ n^a}=log b frac{1}{n^{a-1}}.$
We can assume that $log b ne 0.$
Case 1: $a=1.$ Then $a_n to log b .$
Case 2: $a>1$. Then $a_n to 0.$
Case 3: $a<1.$ Then $a_n to infty$, if $b>1$ and $a_n to - infty$, if $0<b<1.$
answered Feb 1 at 11:29
FredFred
48.5k11849
$endgroup$
Let $a_n:= frac{log b^n}{ n^a}$, then $a_n =frac{n log b}{ n^a}=log b frac{1}{n^{a-1}}.$
We can assume that $log b ne 0.$
Case 1: $a=1.$ Then $a_n to log b .$
Case 2: $a>1$. Then $a_n to 0.$
Case 3: $a<1.$ Then $a_n to infty$, if $b>1$ and $a_n to - infty$, if $0<b<1.$
answered Feb 1 at 11:29
FredFred
48.5k11849
answered Feb 1 at 11:29
FredFred
48.5k11849
answered Feb 1 at 11:29
FredFred
48.5k11849
answered Feb 1 at 11:29
FredFred
48.5k11849
48.5k11849
$begingroup$
My question is, how does this justify replacing the terms (lgn for n and a for 2^a)
$endgroup$
– chakmeshma
Feb 2 at 6:53
add a comment |
$begingroup$
My question is, how does this justify replacing the terms (lgn for n and a for 2^a)
$endgroup$
– chakmeshma
Feb 2 at 6:53
$begingroup$
My question is, how does this justify replacing the terms (lgn for n and a for 2^a)
$endgroup$
– chakmeshma
Feb 2 at 6:53
$begingroup$
My question is, how does this justify replacing the terms (lgn for n and a for 2^a)
$endgroup$
– chakmeshma
Feb 2 at 6:53
add a comment |
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