Mixing
Prove that $f_n$ with $f_n :[0,a]rightarrowmathbb{R},ain mathbb{R}, f_n(x)=x^n$ is continous $forall n in...
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I know how to prove it for $n=2$ can I somehow inductively conclude the Statement $forall ninmathbb{N}$?
Because $forall c,b in[0,a]:L|c-b|geq |c^2-b^2|= |(c+b)(c-b)|iff Lgeq |c+b|iff Lgeq 2a $
real-analysis continuity induction
edited Jan 8 at 21:31
José Carlos Santos
158k22126228
asked Jan 8 at 20:59
RM777RM777
33112
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add a comment |
$begingroup$
I know how to prove it for $n=2$ can I somehow inductively conclude the Statement $forall ninmathbb{N}$?
Because $forall c,b in[0,a]:L|c-b|geq |c^2-b^2|= |(c+b)(c-b)|iff Lgeq |c+b|iff Lgeq 2a $
real-analysis continuity induction
edited Jan 8 at 21:31
José Carlos Santos
158k22126228
asked Jan 8 at 20:59
RM777RM777
33112
$endgroup$
add a comment |
$begingroup$
I know how to prove it for $n=2$ can I somehow inductively conclude the Statement $forall ninmathbb{N}$?
Because $forall c,b in[0,a]:L|c-b|geq |c^2-b^2|= |(c+b)(c-b)|iff Lgeq |c+b|iff Lgeq 2a $
real-analysis continuity induction
edited Jan 8 at 21:31
José Carlos Santos
158k22126228
asked Jan 8 at 20:59
RM777RM777
33112
$endgroup$
I know how to prove it for $n=2$ can I somehow inductively conclude the Statement $forall ninmathbb{N}$?
Because $forall c,b in[0,a]:L|c-b|geq |c^2-b^2|= |(c+b)(c-b)|iff Lgeq |c+b|iff Lgeq 2a $
real-analysis continuity induction
real-analysis continuity induction
edited Jan 8 at 21:31
José Carlos Santos
158k22126228
asked Jan 8 at 20:59
RM777RM777
33112
edited Jan 8 at 21:31
José Carlos Santos
158k22126228
asked Jan 8 at 20:59
RM777RM777
33112
edited Jan 8 at 21:31
José Carlos Santos
158k22126228
edited Jan 8 at 21:31
José Carlos Santos
158k22126228
edited Jan 8 at 21:31
José Carlos Santos
158k22126228
158k22126228
asked Jan 8 at 20:59
RM777RM777
33112
asked Jan 8 at 20:59
RM777RM777
33112
asked Jan 8 at 20:59
RM777RM777
33112
33112
add a comment |
add a comment |
3 Answers
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Yes, you can prove it by induction. If $n=1$, then it is true, because it is the identity function. And if $f_n$ is continuous, then $f_{n+1}$ is continuous too, because $f_{n+1}=f_ntimes f_1$ and the product of two continuous functions is again a continuous function.
answered Jan 8 at 21:04
José Carlos SantosJosé Carlos Santos
158k22126228
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Using the binomial theorem yields:
$$|(c+h)^n-c^n| le sum_{i=1}^n {n choose i} |c|^{n-i}|h|^i xrightarrow{h to 0}$$
because $n in mathbb{N}$ is fixed. Hence $xmapsto x^n$ is continuous at $c$.
answered Jan 8 at 21:14
mechanodroidmechanodroid
27.3k62446
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For any $x,yin [0,a]$ we have
begin{eqnarray}
|x^n-y^n|&=&|x-y|sum_{k=0}^{n-1}x^{n-1-k}y^k\
&le&|x-y|sum_{k=0}^{n-1}x^{n-1-k}y^k\
&le&|x-y|sum_{k=0}^{n-1}a^{n-1}\
&=&na^{n-1}|x-y|
end{eqnarray}
Hence $f_n$ is continuous.
answered Jan 8 at 21:31
Mercy KingMercy King
14k11428
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3 Answers
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active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
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active
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votes
$begingroup$
Yes, you can prove it by induction. If $n=1$, then it is true, because it is the identity function. And if $f_n$ is continuous, then $f_{n+1}$ is continuous too, because $f_{n+1}=f_ntimes f_1$ and the product of two continuous functions is again a continuous function.
answered Jan 8 at 21:04
José Carlos SantosJosé Carlos Santos
158k22126228
$endgroup$
add a comment |
$begingroup$
Yes, you can prove it by induction. If $n=1$, then it is true, because it is the identity function. And if $f_n$ is continuous, then $f_{n+1}$ is continuous too, because $f_{n+1}=f_ntimes f_1$ and the product of two continuous functions is again a continuous function.
answered Jan 8 at 21:04
José Carlos SantosJosé Carlos Santos
158k22126228
$endgroup$
add a comment |
$begingroup$
Yes, you can prove it by induction. If $n=1$, then it is true, because it is the identity function. And if $f_n$ is continuous, then $f_{n+1}$ is continuous too, because $f_{n+1}=f_ntimes f_1$ and the product of two continuous functions is again a continuous function.
answered Jan 8 at 21:04
José Carlos SantosJosé Carlos Santos
158k22126228
$endgroup$
Yes, you can prove it by induction. If $n=1$, then it is true, because it is the identity function. And if $f_n$ is continuous, then $f_{n+1}$ is continuous too, because $f_{n+1}=f_ntimes f_1$ and the product of two continuous functions is again a continuous function.
answered Jan 8 at 21:04
José Carlos SantosJosé Carlos Santos
158k22126228
answered Jan 8 at 21:04
José Carlos SantosJosé Carlos Santos
158k22126228
answered Jan 8 at 21:04
José Carlos SantosJosé Carlos Santos
158k22126228
answered Jan 8 at 21:04
José Carlos SantosJosé Carlos Santos
158k22126228
158k22126228
add a comment |
add a comment |
$begingroup$
Using the binomial theorem yields:
$$|(c+h)^n-c^n| le sum_{i=1}^n {n choose i} |c|^{n-i}|h|^i xrightarrow{h to 0}$$
because $n in mathbb{N}$ is fixed. Hence $xmapsto x^n$ is continuous at $c$.
answered Jan 8 at 21:14
mechanodroidmechanodroid
27.3k62446
$endgroup$
add a comment |
$begingroup$
Using the binomial theorem yields:
$$|(c+h)^n-c^n| le sum_{i=1}^n {n choose i} |c|^{n-i}|h|^i xrightarrow{h to 0}$$
because $n in mathbb{N}$ is fixed. Hence $xmapsto x^n$ is continuous at $c$.
answered Jan 8 at 21:14
mechanodroidmechanodroid
27.3k62446
$endgroup$
add a comment |
$begingroup$
Using the binomial theorem yields:
$$|(c+h)^n-c^n| le sum_{i=1}^n {n choose i} |c|^{n-i}|h|^i xrightarrow{h to 0}$$
because $n in mathbb{N}$ is fixed. Hence $xmapsto x^n$ is continuous at $c$.
answered Jan 8 at 21:14
mechanodroidmechanodroid
27.3k62446
$endgroup$
Using the binomial theorem yields:
$$|(c+h)^n-c^n| le sum_{i=1}^n {n choose i} |c|^{n-i}|h|^i xrightarrow{h to 0}$$
because $n in mathbb{N}$ is fixed. Hence $xmapsto x^n$ is continuous at $c$.
answered Jan 8 at 21:14
mechanodroidmechanodroid
27.3k62446
answered Jan 8 at 21:14
mechanodroidmechanodroid
27.3k62446
answered Jan 8 at 21:14
mechanodroidmechanodroid
27.3k62446
answered Jan 8 at 21:14
mechanodroidmechanodroid
27.3k62446
27.3k62446
add a comment |
add a comment |
$begingroup$
For any $x,yin [0,a]$ we have
begin{eqnarray}
|x^n-y^n|&=&|x-y|sum_{k=0}^{n-1}x^{n-1-k}y^k\
&le&|x-y|sum_{k=0}^{n-1}x^{n-1-k}y^k\
&le&|x-y|sum_{k=0}^{n-1}a^{n-1}\
&=&na^{n-1}|x-y|
end{eqnarray}
Hence $f_n$ is continuous.
answered Jan 8 at 21:31
Mercy KingMercy King
14k11428
$endgroup$
add a comment |
$begingroup$
For any $x,yin [0,a]$ we have
begin{eqnarray}
|x^n-y^n|&=&|x-y|sum_{k=0}^{n-1}x^{n-1-k}y^k\
&le&|x-y|sum_{k=0}^{n-1}x^{n-1-k}y^k\
&le&|x-y|sum_{k=0}^{n-1}a^{n-1}\
&=&na^{n-1}|x-y|
end{eqnarray}
Hence $f_n$ is continuous.
answered Jan 8 at 21:31
Mercy KingMercy King
14k11428
$endgroup$
add a comment |
$begingroup$
For any $x,yin [0,a]$ we have
begin{eqnarray}
|x^n-y^n|&=&|x-y|sum_{k=0}^{n-1}x^{n-1-k}y^k\
&le&|x-y|sum_{k=0}^{n-1}x^{n-1-k}y^k\
&le&|x-y|sum_{k=0}^{n-1}a^{n-1}\
&=&na^{n-1}|x-y|
end{eqnarray}
Hence $f_n$ is continuous.
answered Jan 8 at 21:31
Mercy KingMercy King
14k11428
$endgroup$
For any $x,yin [0,a]$ we have
begin{eqnarray}
|x^n-y^n|&=&|x-y|sum_{k=0}^{n-1}x^{n-1-k}y^k\
&le&|x-y|sum_{k=0}^{n-1}x^{n-1-k}y^k\
&le&|x-y|sum_{k=0}^{n-1}a^{n-1}\
&=&na^{n-1}|x-y|
end{eqnarray}
Hence $f_n$ is continuous.
answered Jan 8 at 21:31
Mercy KingMercy King
14k11428
answered Jan 8 at 21:31
Mercy KingMercy King
14k11428
answered Jan 8 at 21:31
Mercy KingMercy King
14k11428
answered Jan 8 at 21:31
Mercy KingMercy King
14k11428
14k11428
add a comment |
add a comment |
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