What does the graph of $f(x)$ look like if $fbig(f(x)big) = ln(x)$? [closed]
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Best of all I would like the Taylor series of $f(x)$.
real-analysis
edited Aug 27 at 14:28
Bill Wallis
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asked Aug 27 at 13:38
H. Tomasz Grzybowski
523
closed as off-topic by José Carlos Santos, Siong Thye Goh, Adrian Keister, amWhy, Connor Harris Aug 27 at 14:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РJos̩ Carlos Santos, Siong Thye Goh, Adrian Keister, amWhy, Connor Harris
add a comment |Â
up vote
0
down vote
favorite
Best of all I would like the Taylor series of $f(x)$.
real-analysis
edited Aug 27 at 14:28
Bill Wallis
2,2361826
asked Aug 27 at 13:38
H. Tomasz Grzybowski
523
closed as off-topic by José Carlos Santos, Siong Thye Goh, Adrian Keister, amWhy, Connor Harris Aug 27 at 14:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РJos̩ Carlos Santos, Siong Thye Goh, Adrian Keister, amWhy, Connor Harris
Do you want $f$ to be continuous on $(0,infty)$?
– Marco
Aug 27 at 13:55
Similar and recent (but also closed): math.stackexchange.com/questions/2885532/…
– Hans Lundmark
Aug 27 at 14:14
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Best of all I would like the Taylor series of $f(x)$.
real-analysis
edited Aug 27 at 14:28
Bill Wallis
2,2361826
asked Aug 27 at 13:38
H. Tomasz Grzybowski
523
Best of all I would like the Taylor series of $f(x)$.
real-analysis
edited Aug 27 at 14:28
Bill Wallis
2,2361826
asked Aug 27 at 13:38
H. Tomasz Grzybowski
523
edited Aug 27 at 14:28
Bill Wallis
2,2361826
edited Aug 27 at 14:28
Bill Wallis
2,2361826
edited Aug 27 at 14:28
Bill Wallis
2,2361826
2,2361826
asked Aug 27 at 13:38
H. Tomasz Grzybowski
523
asked Aug 27 at 13:38
H. Tomasz Grzybowski
523
asked Aug 27 at 13:38
H. Tomasz Grzybowski
523
523
closed as off-topic by José Carlos Santos, Siong Thye Goh, Adrian Keister, amWhy, Connor Harris Aug 27 at 14:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РJos̩ Carlos Santos, Siong Thye Goh, Adrian Keister, amWhy, Connor Harris
closed as off-topic by José Carlos Santos, Siong Thye Goh, Adrian Keister, amWhy, Connor Harris Aug 27 at 14:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РJos̩ Carlos Santos, Siong Thye Goh, Adrian Keister, amWhy, Connor Harris
Do you want $f$ to be continuous on $(0,infty)$?
– Marco
Aug 27 at 13:55
Similar and recent (but also closed): math.stackexchange.com/questions/2885532/…
– Hans Lundmark
Aug 27 at 14:14
add a comment |Â
Do you want $f$ to be continuous on $(0,infty)$?
– Marco
Aug 27 at 13:55
Similar and recent (but also closed): math.stackexchange.com/questions/2885532/…
– Hans Lundmark
Aug 27 at 14:14
Do you want $f$ to be continuous on $(0,infty)$?
– Marco
Aug 27 at 13:55
Do you want $f$ to be continuous on $(0,infty)$?
– Marco
Aug 27 at 13:55
Similar and recent (but also closed): math.stackexchange.com/questions/2885532/…
– Hans Lundmark
Aug 27 at 14:14
Similar and recent (but also closed): math.stackexchange.com/questions/2885532/…
– Hans Lundmark
Aug 27 at 14:14
add a comment |Â
1 Answer
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Such a continuous function does not exist. First note that $f$ is one-to-one, since if $f(x)=f(y)$, then $f(f(x))=f(f(y))$ and so $x=y$. Then either $f$ is increasing or decreasing on its entire domain $(0,infty)$. Since the range of $ln(x)$ is $mathbbR$, the range of $f$ must be $mathbbR$ as well. If $f$ was decreasing, then $f$ would have to have a fixed point $f(x)=x$ but then $x=f(f(x))=ln(x)$ which is impossible. So $f$ must be increasing on $(0,infty)$ and $lim_x rightarrow 0^+f(x)=-infty$. This is a contradiction, since for $x$ close enough to 0 the quantity $f(f(x))$ would be undefined ($f(x)<0$ there).
answered Aug 27 at 14:01
Marco
1,55917
1
Who said the domain is $(0,infty)$?
– Ivan Neretin
Aug 27 at 14:05
Marco: f(0) = a < 0 and lim f(x) = minus infinity as x -> a. There is an analytic f(.) satisfying f(f(x))= ln(x).
– H. Tomasz Grzybowski
Aug 27 at 16:50
Tomasz, but then $f(f(0))=ln(0)$? The OP is not clear for what $x$ the identity is supposed to hold so I assume the domain is $(0,infty)$.
– Marco
Aug 27 at 17:08
Marco: The domain is (0, infinity). f(f(0)) does not exist, even though f(0) = a, because f(a) does not exist, only the limit as x -> a, f(x) -> minus infinity.
– H. Tomasz Grzybowski
Aug 28 at 20:22
Tomasz, you are right about that, but going back to your previous comment if $f(0)=a$, knowing that $f(x)$ is increasing, this would mean that the range of $f(x)$ is $(a,infty)$. Wouldn't this mean the domain is not $(0,infty)$?
– Marco
Aug 28 at 20:33
 |Â
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Such a continuous function does not exist. First note that $f$ is one-to-one, since if $f(x)=f(y)$, then $f(f(x))=f(f(y))$ and so $x=y$. Then either $f$ is increasing or decreasing on its entire domain $(0,infty)$. Since the range of $ln(x)$ is $mathbbR$, the range of $f$ must be $mathbbR$ as well. If $f$ was decreasing, then $f$ would have to have a fixed point $f(x)=x$ but then $x=f(f(x))=ln(x)$ which is impossible. So $f$ must be increasing on $(0,infty)$ and $lim_x rightarrow 0^+f(x)=-infty$. This is a contradiction, since for $x$ close enough to 0 the quantity $f(f(x))$ would be undefined ($f(x)<0$ there).
answered Aug 27 at 14:01
Marco
1,55917
1
Who said the domain is $(0,infty)$?
– Ivan Neretin
Aug 27 at 14:05
Marco: f(0) = a < 0 and lim f(x) = minus infinity as x -> a. There is an analytic f(.) satisfying f(f(x))= ln(x).
– H. Tomasz Grzybowski
Aug 27 at 16:50
Tomasz, but then $f(f(0))=ln(0)$? The OP is not clear for what $x$ the identity is supposed to hold so I assume the domain is $(0,infty)$.
– Marco
Aug 27 at 17:08
Marco: The domain is (0, infinity). f(f(0)) does not exist, even though f(0) = a, because f(a) does not exist, only the limit as x -> a, f(x) -> minus infinity.
– H. Tomasz Grzybowski
Aug 28 at 20:22
Tomasz, you are right about that, but going back to your previous comment if $f(0)=a$, knowing that $f(x)$ is increasing, this would mean that the range of $f(x)$ is $(a,infty)$. Wouldn't this mean the domain is not $(0,infty)$?
– Marco
Aug 28 at 20:33
 |Â
show 2 more comments
up vote
2
down vote
Such a continuous function does not exist. First note that $f$ is one-to-one, since if $f(x)=f(y)$, then $f(f(x))=f(f(y))$ and so $x=y$. Then either $f$ is increasing or decreasing on its entire domain $(0,infty)$. Since the range of $ln(x)$ is $mathbbR$, the range of $f$ must be $mathbbR$ as well. If $f$ was decreasing, then $f$ would have to have a fixed point $f(x)=x$ but then $x=f(f(x))=ln(x)$ which is impossible. So $f$ must be increasing on $(0,infty)$ and $lim_x rightarrow 0^+f(x)=-infty$. This is a contradiction, since for $x$ close enough to 0 the quantity $f(f(x))$ would be undefined ($f(x)<0$ there).
answered Aug 27 at 14:01
Marco
1,55917
1
Who said the domain is $(0,infty)$?
– Ivan Neretin
Aug 27 at 14:05
Marco: f(0) = a < 0 and lim f(x) = minus infinity as x -> a. There is an analytic f(.) satisfying f(f(x))= ln(x).
– H. Tomasz Grzybowski
Aug 27 at 16:50
Tomasz, but then $f(f(0))=ln(0)$? The OP is not clear for what $x$ the identity is supposed to hold so I assume the domain is $(0,infty)$.
– Marco
Aug 27 at 17:08
Marco: The domain is (0, infinity). f(f(0)) does not exist, even though f(0) = a, because f(a) does not exist, only the limit as x -> a, f(x) -> minus infinity.
– H. Tomasz Grzybowski
Aug 28 at 20:22
Tomasz, you are right about that, but going back to your previous comment if $f(0)=a$, knowing that $f(x)$ is increasing, this would mean that the range of $f(x)$ is $(a,infty)$. Wouldn't this mean the domain is not $(0,infty)$?
– Marco
Aug 28 at 20:33
 |Â
show 2 more comments
up vote
2
down vote
up vote
2
down vote
Such a continuous function does not exist. First note that $f$ is one-to-one, since if $f(x)=f(y)$, then $f(f(x))=f(f(y))$ and so $x=y$. Then either $f$ is increasing or decreasing on its entire domain $(0,infty)$. Since the range of $ln(x)$ is $mathbbR$, the range of $f$ must be $mathbbR$ as well. If $f$ was decreasing, then $f$ would have to have a fixed point $f(x)=x$ but then $x=f(f(x))=ln(x)$ which is impossible. So $f$ must be increasing on $(0,infty)$ and $lim_x rightarrow 0^+f(x)=-infty$. This is a contradiction, since for $x$ close enough to 0 the quantity $f(f(x))$ would be undefined ($f(x)<0$ there).
answered Aug 27 at 14:01
Marco
1,55917
Such a continuous function does not exist. First note that $f$ is one-to-one, since if $f(x)=f(y)$, then $f(f(x))=f(f(y))$ and so $x=y$. Then either $f$ is increasing or decreasing on its entire domain $(0,infty)$. Since the range of $ln(x)$ is $mathbbR$, the range of $f$ must be $mathbbR$ as well. If $f$ was decreasing, then $f$ would have to have a fixed point $f(x)=x$ but then $x=f(f(x))=ln(x)$ which is impossible. So $f$ must be increasing on $(0,infty)$ and $lim_x rightarrow 0^+f(x)=-infty$. This is a contradiction, since for $x$ close enough to 0 the quantity $f(f(x))$ would be undefined ($f(x)<0$ there).
answered Aug 27 at 14:01
Marco
1,55917
answered Aug 27 at 14:01
Marco
1,55917
answered Aug 27 at 14:01
Marco
1,55917
answered Aug 27 at 14:01
Marco
1,55917
1,55917
1
Who said the domain is $(0,infty)$?
– Ivan Neretin
Aug 27 at 14:05
Marco: f(0) = a < 0 and lim f(x) = minus infinity as x -> a. There is an analytic f(.) satisfying f(f(x))= ln(x).
– H. Tomasz Grzybowski
Aug 27 at 16:50
Tomasz, but then $f(f(0))=ln(0)$? The OP is not clear for what $x$ the identity is supposed to hold so I assume the domain is $(0,infty)$.
– Marco
Aug 27 at 17:08
Marco: The domain is (0, infinity). f(f(0)) does not exist, even though f(0) = a, because f(a) does not exist, only the limit as x -> a, f(x) -> minus infinity.
– H. Tomasz Grzybowski
Aug 28 at 20:22
Tomasz, you are right about that, but going back to your previous comment if $f(0)=a$, knowing that $f(x)$ is increasing, this would mean that the range of $f(x)$ is $(a,infty)$. Wouldn't this mean the domain is not $(0,infty)$?
– Marco
Aug 28 at 20:33
 |Â
show 2 more comments
1
Who said the domain is $(0,infty)$?
– Ivan Neretin
Aug 27 at 14:05
Marco: f(0) = a < 0 and lim f(x) = minus infinity as x -> a. There is an analytic f(.) satisfying f(f(x))= ln(x).
– H. Tomasz Grzybowski
Aug 27 at 16:50
Tomasz, but then $f(f(0))=ln(0)$? The OP is not clear for what $x$ the identity is supposed to hold so I assume the domain is $(0,infty)$.
– Marco
Aug 27 at 17:08
Marco: The domain is (0, infinity). f(f(0)) does not exist, even though f(0) = a, because f(a) does not exist, only the limit as x -> a, f(x) -> minus infinity.
– H. Tomasz Grzybowski
Aug 28 at 20:22
Tomasz, you are right about that, but going back to your previous comment if $f(0)=a$, knowing that $f(x)$ is increasing, this would mean that the range of $f(x)$ is $(a,infty)$. Wouldn't this mean the domain is not $(0,infty)$?
– Marco
Aug 28 at 20:33
1
1
Who said the domain is $(0,infty)$?
– Ivan Neretin
Aug 27 at 14:05
Who said the domain is $(0,infty)$?
– Ivan Neretin
Aug 27 at 14:05
Marco: f(0) = a < 0 and lim f(x) = minus infinity as x -> a. There is an analytic f(.) satisfying f(f(x))= ln(x).
– H. Tomasz Grzybowski
Aug 27 at 16:50
Marco: f(0) = a < 0 and lim f(x) = minus infinity as x -> a. There is an analytic f(.) satisfying f(f(x))= ln(x).
– H. Tomasz Grzybowski
Aug 27 at 16:50
Tomasz, but then $f(f(0))=ln(0)$? The OP is not clear for what $x$ the identity is supposed to hold so I assume the domain is $(0,infty)$.
– Marco
Aug 27 at 17:08
Tomasz, but then $f(f(0))=ln(0)$? The OP is not clear for what $x$ the identity is supposed to hold so I assume the domain is $(0,infty)$.
– Marco
Aug 27 at 17:08
Marco: The domain is (0, infinity). f(f(0)) does not exist, even though f(0) = a, because f(a) does not exist, only the limit as x -> a, f(x) -> minus infinity.
– H. Tomasz Grzybowski
Aug 28 at 20:22
Marco: The domain is (0, infinity). f(f(0)) does not exist, even though f(0) = a, because f(a) does not exist, only the limit as x -> a, f(x) -> minus infinity.
– H. Tomasz Grzybowski
Aug 28 at 20:22
Tomasz, you are right about that, but going back to your previous comment if $f(0)=a$, knowing that $f(x)$ is increasing, this would mean that the range of $f(x)$ is $(a,infty)$. Wouldn't this mean the domain is not $(0,infty)$?
– Marco
Aug 28 at 20:33
Tomasz, you are right about that, but going back to your previous comment if $f(0)=a$, knowing that $f(x)$ is increasing, this would mean that the range of $f(x)$ is $(a,infty)$. Wouldn't this mean the domain is not $(0,infty)$?
– Marco
Aug 28 at 20:33
 |Â
show 2 more comments
Do you want $f$ to be continuous on $(0,infty)$?
– Marco
Aug 27 at 13:55
Similar and recent (but also closed): math.stackexchange.com/questions/2885532/…
– Hans Lundmark
Aug 27 at 14:14