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Show voter details
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moderate
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Proxies\__CG__\App\Entity\Entry {#1551
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+title: "A meme for math people"
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It's been a while, but I think I remember this one. Lim 1/n =0 as n approaches infinity. Let x^0 be undefined. For any e>0 there exists an n such that |x^(1/n) -1| < e. If you desire x^(1/n) to be continuous at 0, you define x^0 as 1.\n
\n
E2a: since x^(1/n)>1, you can drop the abs bars. I think you can get an inequality to pick n using logs.
"""
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…2
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Show voter details
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3 |
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edit
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Proxies\__CG__\App\Entity\Entry {#1551
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It's been a while, but I think I remember this one. Lim 1/n =0 as n approaches infinity. Let x^0 be undefined. For any e>0 there exists an n such that |x^(1/n) -1| < e. If you desire x^(1/n) to be continuous at 0, you define x^0 as 1.\n
\n
E2a: since x^(1/n)>1, you can drop the abs bars. I think you can get an inequality to pick n using logs.
"""
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…2
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Show voter details
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4 |
DENIED
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moderate
|
Proxies\__CG__\App\Entity\Entry {#1551
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It's been a while, but I think I remember this one. Lim 1/n =0 as n approaches infinity. Let x^0 be undefined. For any e>0 there exists an n such that |x^(1/n) -1| < e. If you desire x^(1/n) to be continuous at 0, you define x^0 as 1.\n
\n
E2a: since x^(1/n)>1, you can drop the abs bars. I think you can get an inequality to pick n using logs.
"""
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…2
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Show voter details
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5 |
DENIED
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ROLE_USER
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null |
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Show voter details
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6 |
DENIED
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moderate
|
App\Entity\EntryComment {#1676
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It's been a while, but I think I remember this one. Lim 1/n =0 as n approaches infinity. Let x^0 be undefined. For any e>0 there exists an n such that |x^(1/n) -1| < e. If you desire x^(1/n) to be continuous at 0, you define x^0 as 1.\n
\n
E2a: since x^(1/n)>1, you can drop the abs bars. I think you can get an inequality to pick n using logs.
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Show voter details
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7 |
DENIED
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edit
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App\Entity\EntryComment {#1676
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…2
}
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It's been a while, but I think I remember this one. Lim 1/n =0 as n approaches infinity. Let x^0 be undefined. For any e>0 there exists an n such that |x^(1/n) -1| < e. If you desire x^(1/n) to be continuous at 0, you define x^0 as 1.\n
\n
E2a: since x^(1/n)>1, you can drop the abs bars. I think you can get an inequality to pick n using logs.
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Show voter details
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8 |
DENIED
|
moderate
|
App\Entity\EntryComment {#1676
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It's been a while, but I think I remember this one. Lim 1/n =0 as n approaches infinity. Let x^0 be undefined. For any e>0 there exists an n such that |x^(1/n) -1| < e. If you desire x^(1/n) to be continuous at 0, you define x^0 as 1.\n
\n
E2a: since x^(1/n)>1, you can drop the abs bars. I think you can get an inequality to pick n using logs.
"""
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