Find lower bound of function ？

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Can someone help me find a lower bound to the function



$$ln(1 - 0.5^u), uge0$$



I tried to use the Taylor formula, but it didn't seem to work. I would be grateful if anyone could give me some advice.




calculus functions optimization




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  • 1




    I guess you wanted to write $u>0$, as the expression is not defined at $0$.
    – A. Pongrácz
    Aug 12 at 8:24














up vote
0
down vote

favorite












Can someone help me find a lower bound to the function



$$ln(1 - 0.5^u), uge0$$



I tried to use the Taylor formula, but it didn't seem to work. I would be grateful if anyone could give me some advice.




calculus functions optimization




share|cite|improve this question


















  • 1




    I guess you wanted to write $u>0$, as the expression is not defined at $0$.
    – A. Pongrácz
    Aug 12 at 8:24












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Can someone help me find a lower bound to the function



$$ln(1 - 0.5^u), uge0$$



I tried to use the Taylor formula, but it didn't seem to work. I would be grateful if anyone could give me some advice.




calculus functions optimization




share|cite|improve this question














Can someone help me find a lower bound to the function



$$ln(1 - 0.5^u), uge0$$



I tried to use the Taylor formula, but it didn't seem to work. I would be grateful if anyone could give me some advice.





calculus functions optimization





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 12 at 8:20









Adrian

5,1991035




5,1991035










asked Aug 12 at 8:16









J. Doe

1




1







  • 1




    I guess you wanted to write $u>0$, as the expression is not defined at $0$.
    – A. Pongrácz
    Aug 12 at 8:24












  • 1




    I guess you wanted to write $u>0$, as the expression is not defined at $0$.
    – A. Pongrácz
    Aug 12 at 8:24







1




1




I guess you wanted to write $u>0$, as the expression is not defined at $0$.
– A. Pongrácz
Aug 12 at 8:24




I guess you wanted to write $u>0$, as the expression is not defined at $0$.
– A. Pongrácz
Aug 12 at 8:24










2 Answers
2






active

oldest

votes

















up vote
2
down vote













As $urightarrow 0$, the expression tends to $-infty$, so there is no lower bound.






share|cite|improve this answer




















  • Thank for your reply. if $u>=1$?
    – J. Doe
    Aug 12 at 8:39






  • 1




    The function is clerly strictly monotone increasing. So in that case, the minimum is $f(1)$.
    – A. Pongrácz
    Aug 12 at 8:44










  • Thanks a lot. but I want to get a lower bound function, not to get minimum. Could you give me some advice?
    – J. Doe
    Aug 12 at 10:13











  • You asked lower bound. I also don't understand the difference. $f(1)=ln 0.5 approx -0.7$ is a global lower bound. If you want, this can be rephrased as: the constant $-0.7$ function is below $f$ everywhere.
    – A. Pongrácz
    Aug 12 at 10:19











  • yeah, constant $-0.7$ is ok, but it is a constant function. I want to get a non constant function.
    – J. Doe
    Aug 12 at 10:38

















up vote
2
down vote













As a general approach you could try to find a minimum point.



$$y=ln(1-0.5^u).$$



Differentiating,



$$y'=fraclog(2)0.5^u1-u^0.5.$$



To find stationary points, set equal to zero to obtain



$$0.5^u=0,$$



which has no solutions, so there are no stationary points. Hence, no minimum and so no lower bound.






share|cite|improve this answer






















  • Thanks a lot. but my purpose is to get a lower bound function, not to get minimum.
    – J. Doe
    Aug 12 at 10:02










  • $f(u)=c$ where $c$ is constant is a lower bound function. Maybe you mean a non constant function..
    – Antinous
    Aug 12 at 10:19











  • You just keep changing the problem. Be decisive. Also, what purpose would such an answer serve? I can easlily give you a linear lower bound (why is that any better than a constant lower bound, which was your original problem, is beyond me, though), but then you will say you really want to get a non-constant functionwhose value at 7.63 is 4.12 and that is not continuous at $pi$. What is the motivation of the problem, and what is the answer you are really looking for? As soon as you can answer these questions, maybe others will help you. Also, I do not see any of your own efforts.
    – A. Pongrácz
    Aug 12 at 10:43











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote













As $urightarrow 0$, the expression tends to $-infty$, so there is no lower bound.






share|cite|improve this answer




















  • Thank for your reply. if $u>=1$?
    – J. Doe
    Aug 12 at 8:39






  • 1




    The function is clerly strictly monotone increasing. So in that case, the minimum is $f(1)$.
    – A. Pongrácz
    Aug 12 at 8:44










  • Thanks a lot. but I want to get a lower bound function, not to get minimum. Could you give me some advice?
    – J. Doe
    Aug 12 at 10:13











  • You asked lower bound. I also don't understand the difference. $f(1)=ln 0.5 approx -0.7$ is a global lower bound. If you want, this can be rephrased as: the constant $-0.7$ function is below $f$ everywhere.
    – A. Pongrácz
    Aug 12 at 10:19











  • yeah, constant $-0.7$ is ok, but it is a constant function. I want to get a non constant function.
    – J. Doe
    Aug 12 at 10:38














up vote
2
down vote













As $urightarrow 0$, the expression tends to $-infty$, so there is no lower bound.






share|cite|improve this answer




















  • Thank for your reply. if $u>=1$?
    – J. Doe
    Aug 12 at 8:39






  • 1




    The function is clerly strictly monotone increasing. So in that case, the minimum is $f(1)$.
    – A. Pongrácz
    Aug 12 at 8:44










  • Thanks a lot. but I want to get a lower bound function, not to get minimum. Could you give me some advice?
    – J. Doe
    Aug 12 at 10:13











  • You asked lower bound. I also don't understand the difference. $f(1)=ln 0.5 approx -0.7$ is a global lower bound. If you want, this can be rephrased as: the constant $-0.7$ function is below $f$ everywhere.
    – A. Pongrácz
    Aug 12 at 10:19











  • yeah, constant $-0.7$ is ok, but it is a constant function. I want to get a non constant function.
    – J. Doe
    Aug 12 at 10:38












up vote
2
down vote










up vote
2
down vote









As $urightarrow 0$, the expression tends to $-infty$, so there is no lower bound.






share|cite|improve this answer












As $urightarrow 0$, the expression tends to $-infty$, so there is no lower bound.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 12 at 8:26









A. Pongrácz

3,677624




3,677624











  • Thank for your reply. if $u>=1$?
    – J. Doe
    Aug 12 at 8:39






  • 1




    The function is clerly strictly monotone increasing. So in that case, the minimum is $f(1)$.
    – A. Pongrácz
    Aug 12 at 8:44










  • Thanks a lot. but I want to get a lower bound function, not to get minimum. Could you give me some advice?
    – J. Doe
    Aug 12 at 10:13











  • You asked lower bound. I also don't understand the difference. $f(1)=ln 0.5 approx -0.7$ is a global lower bound. If you want, this can be rephrased as: the constant $-0.7$ function is below $f$ everywhere.
    – A. Pongrácz
    Aug 12 at 10:19











  • yeah, constant $-0.7$ is ok, but it is a constant function. I want to get a non constant function.
    – J. Doe
    Aug 12 at 10:38
















  • Thank for your reply. if $u>=1$?
    – J. Doe
    Aug 12 at 8:39






  • 1




    The function is clerly strictly monotone increasing. So in that case, the minimum is $f(1)$.
    – A. Pongrácz
    Aug 12 at 8:44










  • Thanks a lot. but I want to get a lower bound function, not to get minimum. Could you give me some advice?
    – J. Doe
    Aug 12 at 10:13











  • You asked lower bound. I also don't understand the difference. $f(1)=ln 0.5 approx -0.7$ is a global lower bound. If you want, this can be rephrased as: the constant $-0.7$ function is below $f$ everywhere.
    – A. Pongrácz
    Aug 12 at 10:19











  • yeah, constant $-0.7$ is ok, but it is a constant function. I want to get a non constant function.
    – J. Doe
    Aug 12 at 10:38















Thank for your reply. if $u>=1$?
– J. Doe
Aug 12 at 8:39




Thank for your reply. if $u>=1$?
– J. Doe
Aug 12 at 8:39




1




1




The function is clerly strictly monotone increasing. So in that case, the minimum is $f(1)$.
– A. Pongrácz
Aug 12 at 8:44




The function is clerly strictly monotone increasing. So in that case, the minimum is $f(1)$.
– A. Pongrácz
Aug 12 at 8:44












Thanks a lot. but I want to get a lower bound function, not to get minimum. Could you give me some advice?
– J. Doe
Aug 12 at 10:13





Thanks a lot. but I want to get a lower bound function, not to get minimum. Could you give me some advice?
– J. Doe
Aug 12 at 10:13













You asked lower bound. I also don't understand the difference. $f(1)=ln 0.5 approx -0.7$ is a global lower bound. If you want, this can be rephrased as: the constant $-0.7$ function is below $f$ everywhere.
– A. Pongrácz
Aug 12 at 10:19





You asked lower bound. I also don't understand the difference. $f(1)=ln 0.5 approx -0.7$ is a global lower bound. If you want, this can be rephrased as: the constant $-0.7$ function is below $f$ everywhere.
– A. Pongrácz
Aug 12 at 10:19













yeah, constant $-0.7$ is ok, but it is a constant function. I want to get a non constant function.
– J. Doe
Aug 12 at 10:38




yeah, constant $-0.7$ is ok, but it is a constant function. I want to get a non constant function.
– J. Doe
Aug 12 at 10:38










up vote
2
down vote













As a general approach you could try to find a minimum point.



$$y=ln(1-0.5^u).$$



Differentiating,



$$y'=fraclog(2)0.5^u1-u^0.5.$$



To find stationary points, set equal to zero to obtain



$$0.5^u=0,$$



which has no solutions, so there are no stationary points. Hence, no minimum and so no lower bound.






share|cite|improve this answer






















  • Thanks a lot. but my purpose is to get a lower bound function, not to get minimum.
    – J. Doe
    Aug 12 at 10:02










  • $f(u)=c$ where $c$ is constant is a lower bound function. Maybe you mean a non constant function..
    – Antinous
    Aug 12 at 10:19











  • You just keep changing the problem. Be decisive. Also, what purpose would such an answer serve? I can easlily give you a linear lower bound (why is that any better than a constant lower bound, which was your original problem, is beyond me, though), but then you will say you really want to get a non-constant functionwhose value at 7.63 is 4.12 and that is not continuous at $pi$. What is the motivation of the problem, and what is the answer you are really looking for? As soon as you can answer these questions, maybe others will help you. Also, I do not see any of your own efforts.
    – A. Pongrácz
    Aug 12 at 10:43















up vote
2
down vote













As a general approach you could try to find a minimum point.



$$y=ln(1-0.5^u).$$



Differentiating,



$$y'=fraclog(2)0.5^u1-u^0.5.$$



To find stationary points, set equal to zero to obtain



$$0.5^u=0,$$



which has no solutions, so there are no stationary points. Hence, no minimum and so no lower bound.






share|cite|improve this answer






















  • Thanks a lot. but my purpose is to get a lower bound function, not to get minimum.
    – J. Doe
    Aug 12 at 10:02










  • $f(u)=c$ where $c$ is constant is a lower bound function. Maybe you mean a non constant function..
    – Antinous
    Aug 12 at 10:19











  • You just keep changing the problem. Be decisive. Also, what purpose would such an answer serve? I can easlily give you a linear lower bound (why is that any better than a constant lower bound, which was your original problem, is beyond me, though), but then you will say you really want to get a non-constant functionwhose value at 7.63 is 4.12 and that is not continuous at $pi$. What is the motivation of the problem, and what is the answer you are really looking for? As soon as you can answer these questions, maybe others will help you. Also, I do not see any of your own efforts.
    – A. Pongrácz
    Aug 12 at 10:43













up vote
2
down vote










up vote
2
down vote









As a general approach you could try to find a minimum point.



$$y=ln(1-0.5^u).$$



Differentiating,



$$y'=fraclog(2)0.5^u1-u^0.5.$$



To find stationary points, set equal to zero to obtain



$$0.5^u=0,$$



which has no solutions, so there are no stationary points. Hence, no minimum and so no lower bound.






share|cite|improve this answer














As a general approach you could try to find a minimum point.



$$y=ln(1-0.5^u).$$



Differentiating,



$$y'=fraclog(2)0.5^u1-u^0.5.$$



To find stationary points, set equal to zero to obtain



$$0.5^u=0,$$



which has no solutions, so there are no stationary points. Hence, no minimum and so no lower bound.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Aug 12 at 8:50

























answered Aug 12 at 8:43









Antinous

5,47541950




5,47541950











  • Thanks a lot. but my purpose is to get a lower bound function, not to get minimum.
    – J. Doe
    Aug 12 at 10:02










  • $f(u)=c$ where $c$ is constant is a lower bound function. Maybe you mean a non constant function..
    – Antinous
    Aug 12 at 10:19











  • You just keep changing the problem. Be decisive. Also, what purpose would such an answer serve? I can easlily give you a linear lower bound (why is that any better than a constant lower bound, which was your original problem, is beyond me, though), but then you will say you really want to get a non-constant functionwhose value at 7.63 is 4.12 and that is not continuous at $pi$. What is the motivation of the problem, and what is the answer you are really looking for? As soon as you can answer these questions, maybe others will help you. Also, I do not see any of your own efforts.
    – A. Pongrácz
    Aug 12 at 10:43

















  • Thanks a lot. but my purpose is to get a lower bound function, not to get minimum.
    – J. Doe
    Aug 12 at 10:02










  • $f(u)=c$ where $c$ is constant is a lower bound function. Maybe you mean a non constant function..
    – Antinous
    Aug 12 at 10:19











  • You just keep changing the problem. Be decisive. Also, what purpose would such an answer serve? I can easlily give you a linear lower bound (why is that any better than a constant lower bound, which was your original problem, is beyond me, though), but then you will say you really want to get a non-constant functionwhose value at 7.63 is 4.12 and that is not continuous at $pi$. What is the motivation of the problem, and what is the answer you are really looking for? As soon as you can answer these questions, maybe others will help you. Also, I do not see any of your own efforts.
    – A. Pongrácz
    Aug 12 at 10:43
















Thanks a lot. but my purpose is to get a lower bound function, not to get minimum.
– J. Doe
Aug 12 at 10:02




Thanks a lot. but my purpose is to get a lower bound function, not to get minimum.
– J. Doe
Aug 12 at 10:02












$f(u)=c$ where $c$ is constant is a lower bound function. Maybe you mean a non constant function..
– Antinous
Aug 12 at 10:19





$f(u)=c$ where $c$ is constant is a lower bound function. Maybe you mean a non constant function..
– Antinous
Aug 12 at 10:19













You just keep changing the problem. Be decisive. Also, what purpose would such an answer serve? I can easlily give you a linear lower bound (why is that any better than a constant lower bound, which was your original problem, is beyond me, though), but then you will say you really want to get a non-constant functionwhose value at 7.63 is 4.12 and that is not continuous at $pi$. What is the motivation of the problem, and what is the answer you are really looking for? As soon as you can answer these questions, maybe others will help you. Also, I do not see any of your own efforts.
– A. Pongrácz
Aug 12 at 10:43





You just keep changing the problem. Be decisive. Also, what purpose would such an answer serve? I can easlily give you a linear lower bound (why is that any better than a constant lower bound, which was your original problem, is beyond me, though), but then you will say you really want to get a non-constant functionwhose value at 7.63 is 4.12 and that is not continuous at $pi$. What is the motivation of the problem, and what is the answer you are really looking for? As soon as you can answer these questions, maybe others will help you. Also, I do not see any of your own efforts.
– A. Pongrácz
Aug 12 at 10:43













 

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