Prove that the function $f:mathbbRtomathbbR$ by $f(x)=x^frac1n$ is continuous
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Prove that the function $f:mathbbRtomathbbR$ by $f(x)=x^frac1p$ is continuous
Here, I am trying to prove by sequential criterion.
Suppose $f:(0,infty)to mathbbR$ by $f(x)=x^frac1p$
I need to prove that if $x_nto a$ then $f(x_n)to f(a)$
I am choosing $y_n=f(x_n)=x_n^frac1p$ and $b=f(a)=a^frac1n$
Now, I consider $y_n^p-b^p=(y_n-b)(y_n^p-1+........+b^p-1)$
$Rightarrow |y_n-b|le fracy_n^p-b^p\Rightarrow |f(x_n)-f(a)|le fracx_n-ab^p-1<epsilon$
as $x_nto a$
So, I proved for positive real numbers. How can I prove for negative real numbers? For example, for $x=-1$?
real-analysis limits continuity
edited Sep 6 at 7:14
GoodDeeds
10.2k21335
asked Sep 6 at 7:11
Inverse Problem
906918
add a comment |Â
up vote
1
down vote
favorite
Prove that the function $f:mathbbRtomathbbR$ by $f(x)=x^frac1p$ is continuous
Here, I am trying to prove by sequential criterion.
Suppose $f:(0,infty)to mathbbR$ by $f(x)=x^frac1p$
I need to prove that if $x_nto a$ then $f(x_n)to f(a)$
I am choosing $y_n=f(x_n)=x_n^frac1p$ and $b=f(a)=a^frac1n$
Now, I consider $y_n^p-b^p=(y_n-b)(y_n^p-1+........+b^p-1)$
$Rightarrow |y_n-b|le fracy_n^p-b^p\Rightarrow |f(x_n)-f(a)|le fracx_n-ab^p-1<epsilon$
as $x_nto a$
So, I proved for positive real numbers. How can I prove for negative real numbers? For example, for $x=-1$?
real-analysis limits continuity
edited Sep 6 at 7:14
GoodDeeds
10.2k21335
asked Sep 6 at 7:11
Inverse Problem
906918
What is $p$? If you take $p=2$ then $f$ is not even defined as a function that maps $mathbbR$ into $mathbbR$.
– Sobi
Sep 6 at 7:14
@sobi ...we solve this problem with odd and even
– Inverse Problem
Sep 6 at 7:15
@sobi..natural number
– Inverse Problem
Sep 6 at 7:15
OP: if the domain depends on $p$, then you need to separate these cases out. As @Sobi states, if $p=2$, then the domain can’t be $Bbb R$, though it could be $Bbb R^+$ or similar.
– Clayton
Sep 6 at 7:23
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Prove that the function $f:mathbbRtomathbbR$ by $f(x)=x^frac1p$ is continuous
Here, I am trying to prove by sequential criterion.
Suppose $f:(0,infty)to mathbbR$ by $f(x)=x^frac1p$
I need to prove that if $x_nto a$ then $f(x_n)to f(a)$
I am choosing $y_n=f(x_n)=x_n^frac1p$ and $b=f(a)=a^frac1n$
Now, I consider $y_n^p-b^p=(y_n-b)(y_n^p-1+........+b^p-1)$
$Rightarrow |y_n-b|le fracy_n^p-b^p\Rightarrow |f(x_n)-f(a)|le fracx_n-ab^p-1<epsilon$
as $x_nto a$
So, I proved for positive real numbers. How can I prove for negative real numbers? For example, for $x=-1$?
real-analysis limits continuity
edited Sep 6 at 7:14
GoodDeeds
10.2k21335
asked Sep 6 at 7:11
Inverse Problem
906918
Prove that the function $f:mathbbRtomathbbR$ by $f(x)=x^frac1p$ is continuous
Here, I am trying to prove by sequential criterion.
Suppose $f:(0,infty)to mathbbR$ by $f(x)=x^frac1p$
I need to prove that if $x_nto a$ then $f(x_n)to f(a)$
I am choosing $y_n=f(x_n)=x_n^frac1p$ and $b=f(a)=a^frac1n$
Now, I consider $y_n^p-b^p=(y_n-b)(y_n^p-1+........+b^p-1)$
$Rightarrow |y_n-b|le fracy_n^p-b^p\Rightarrow |f(x_n)-f(a)|le fracx_n-ab^p-1<epsilon$
as $x_nto a$
So, I proved for positive real numbers. How can I prove for negative real numbers? For example, for $x=-1$?
real-analysis limits continuity
real-analysis limits continuity
edited Sep 6 at 7:14
GoodDeeds
10.2k21335
asked Sep 6 at 7:11
Inverse Problem
906918
edited Sep 6 at 7:14
GoodDeeds
10.2k21335
asked Sep 6 at 7:11
Inverse Problem
906918
edited Sep 6 at 7:14
GoodDeeds
10.2k21335
edited Sep 6 at 7:14
GoodDeeds
10.2k21335
edited Sep 6 at 7:14
GoodDeeds
10.2k21335
10.2k21335
asked Sep 6 at 7:11
Inverse Problem
906918
asked Sep 6 at 7:11
Inverse Problem
906918
asked Sep 6 at 7:11
Inverse Problem
906918
906918
What is $p$? If you take $p=2$ then $f$ is not even defined as a function that maps $mathbbR$ into $mathbbR$.
– Sobi
Sep 6 at 7:14
@sobi ...we solve this problem with odd and even
– Inverse Problem
Sep 6 at 7:15
@sobi..natural number
– Inverse Problem
Sep 6 at 7:15
OP: if the domain depends on $p$, then you need to separate these cases out. As @Sobi states, if $p=2$, then the domain can’t be $Bbb R$, though it could be $Bbb R^+$ or similar.
– Clayton
Sep 6 at 7:23
add a comment |Â
What is $p$? If you take $p=2$ then $f$ is not even defined as a function that maps $mathbbR$ into $mathbbR$.
– Sobi
Sep 6 at 7:14
@sobi ...we solve this problem with odd and even
– Inverse Problem
Sep 6 at 7:15
@sobi..natural number
– Inverse Problem
Sep 6 at 7:15
OP: if the domain depends on $p$, then you need to separate these cases out. As @Sobi states, if $p=2$, then the domain can’t be $Bbb R$, though it could be $Bbb R^+$ or similar.
– Clayton
Sep 6 at 7:23
What is $p$? If you take $p=2$ then $f$ is not even defined as a function that maps $mathbbR$ into $mathbbR$.
– Sobi
Sep 6 at 7:14
What is $p$? If you take $p=2$ then $f$ is not even defined as a function that maps $mathbbR$ into $mathbbR$.
– Sobi
Sep 6 at 7:14
@sobi ...we solve this problem with odd and even
– Inverse Problem
Sep 6 at 7:15
@sobi ...we solve this problem with odd and even
– Inverse Problem
Sep 6 at 7:15
@sobi..natural number
– Inverse Problem
Sep 6 at 7:15
@sobi..natural number
– Inverse Problem
Sep 6 at 7:15
OP: if the domain depends on $p$, then you need to separate these cases out. As @Sobi states, if $p=2$, then the domain can’t be $Bbb R$, though it could be $Bbb R^+$ or similar.
– Clayton
Sep 6 at 7:23
OP: if the domain depends on $p$, then you need to separate these cases out. As @Sobi states, if $p=2$, then the domain can’t be $Bbb R$, though it could be $Bbb R^+$ or similar.
– Clayton
Sep 6 at 7:23
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
The function $f(x)=sqrt[n]x$ only is defined on $[0,infty)$ for even $n$. For odd $n$, write $f(x)=sqrt[n]x$ where $x>0$ and $f(x)=-sqrt[n]-x$ where $x<0$.
answered Sep 6 at 7:25
Nosrati
22.5k61748
@nosrati....we prove for x<0 then n must be odd right
– Inverse Problem
Sep 6 at 7:31
for odd $n$, when $x<0$ you can use $-sqrt[n]-x$ for $-x>0$.
– Nosrati
Sep 6 at 7:33
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The function $f(x)=sqrt[n]x$ only is defined on $[0,infty)$ for even $n$. For odd $n$, write $f(x)=sqrt[n]x$ where $x>0$ and $f(x)=-sqrt[n]-x$ where $x<0$.
answered Sep 6 at 7:25
Nosrati
22.5k61748
@nosrati....we prove for x<0 then n must be odd right
– Inverse Problem
Sep 6 at 7:31
for odd $n$, when $x<0$ you can use $-sqrt[n]-x$ for $-x>0$.
– Nosrati
Sep 6 at 7:33
add a comment |Â
up vote
2
down vote
accepted
The function $f(x)=sqrt[n]x$ only is defined on $[0,infty)$ for even $n$. For odd $n$, write $f(x)=sqrt[n]x$ where $x>0$ and $f(x)=-sqrt[n]-x$ where $x<0$.
answered Sep 6 at 7:25
Nosrati
22.5k61748
@nosrati....we prove for x<0 then n must be odd right
– Inverse Problem
Sep 6 at 7:31
for odd $n$, when $x<0$ you can use $-sqrt[n]-x$ for $-x>0$.
– Nosrati
Sep 6 at 7:33
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The function $f(x)=sqrt[n]x$ only is defined on $[0,infty)$ for even $n$. For odd $n$, write $f(x)=sqrt[n]x$ where $x>0$ and $f(x)=-sqrt[n]-x$ where $x<0$.
answered Sep 6 at 7:25
Nosrati
22.5k61748
The function $f(x)=sqrt[n]x$ only is defined on $[0,infty)$ for even $n$. For odd $n$, write $f(x)=sqrt[n]x$ where $x>0$ and $f(x)=-sqrt[n]-x$ where $x<0$.
answered Sep 6 at 7:25
Nosrati
22.5k61748
answered Sep 6 at 7:25
Nosrati
22.5k61748
answered Sep 6 at 7:25
Nosrati
22.5k61748
answered Sep 6 at 7:25
Nosrati
22.5k61748
22.5k61748
@nosrati....we prove for x<0 then n must be odd right
– Inverse Problem
Sep 6 at 7:31
for odd $n$, when $x<0$ you can use $-sqrt[n]-x$ for $-x>0$.
– Nosrati
Sep 6 at 7:33
add a comment |Â
@nosrati....we prove for x<0 then n must be odd right
– Inverse Problem
Sep 6 at 7:31
for odd $n$, when $x<0$ you can use $-sqrt[n]-x$ for $-x>0$.
– Nosrati
Sep 6 at 7:33
@nosrati....we prove for x<0 then n must be odd right
– Inverse Problem
Sep 6 at 7:31
@nosrati....we prove for x<0 then n must be odd right
– Inverse Problem
Sep 6 at 7:31
for odd $n$, when $x<0$ you can use $-sqrt[n]-x$ for $-x>0$.
– Nosrati
Sep 6 at 7:33
for odd $n$, when $x<0$ you can use $-sqrt[n]-x$ for $-x>0$.
– Nosrati
Sep 6 at 7:33
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2907180%2fprove-that-the-function-f-mathbbr-to-mathbbr-by-fx-x-frac1n-is-c%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
What is $p$? If you take $p=2$ then $f$ is not even defined as a function that maps $mathbbR$ into $mathbbR$.
– Sobi
Sep 6 at 7:14
@sobi ...we solve this problem with odd and even
– Inverse Problem
Sep 6 at 7:15
@sobi..natural number
– Inverse Problem
Sep 6 at 7:15
OP: if the domain depends on $p$, then you need to separate these cases out. As @Sobi states, if $p=2$, then the domain can’t be $Bbb R$, though it could be $Bbb R^+$ or similar.
– Clayton
Sep 6 at 7:23