Will this inequality stay preserved if I take $log$ on both sides? EDIT: Also, can someone please try and...











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1
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I am trying to solved this inequality for $k$.



$x^{2k}<varepsiloncdot k^k$



Here $kinmathbb{N}$ and $x,varepsilon$ are fixed such that $x,varepsiloninmathbb{R}$ and $varepsilon>0$. I was thinking about taking $log_varepsilon$ on both sides but I am not sure about whether this will preserve the inequality or not? Moreover, I cannot really think of any other way to isolate $k$ other than to consider logarithms.



EDIT: Can someone suggest a solution?



Also if possible, could someone tell me what are the conditions for when the inequalities are preserved under logarithms and what are the properties that need to be satisfied by the base? EDIT: Got the answer for this! Thank you!










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  • 4




    It's possible to show that if the function you're using is strictly monotonically increasing (which the logarithm is), it will preserve inequalities. If it's strictly monotonically decreasing, it will reverse the inequality. If it's neither, you will destroy the inequality.
    – Adrian Keister
    Nov 14 at 19:41










  • If $varepsilon>1$, the inequality will be preserved— but if $0<varepsilon<1$, the inequality will be reversed.
    – Mercy King
    Nov 14 at 19:47










  • @AdrianKeister and Mercy King : Thank you for your comments. So does that mean the value of $x$ makes no difference and does not have an impact on how the inequality will be upon taking the logarithm? EDIT : Also, what about $varepsilon=1$? Do you just deal with that as a special case?
    – Bhavesh Singhal
    Nov 14 at 19:59












  • Rather than take $log_{varepsilon},$ I would just do $ln$ of both sides.
    – Adrian Keister
    Nov 14 at 20:12










  • @MercyKing: I think the conditions on the variables present would make that an impossibility, right? The LHS is non-negative, and $k^k>0$ for all $kinmathbb{N}$, so $varepsilon<0$ would make the inequality a contradiction.
    – Adrian Keister
    Nov 14 at 20:14















up vote
1
down vote

favorite












I am trying to solved this inequality for $k$.



$x^{2k}<varepsiloncdot k^k$



Here $kinmathbb{N}$ and $x,varepsilon$ are fixed such that $x,varepsiloninmathbb{R}$ and $varepsilon>0$. I was thinking about taking $log_varepsilon$ on both sides but I am not sure about whether this will preserve the inequality or not? Moreover, I cannot really think of any other way to isolate $k$ other than to consider logarithms.



EDIT: Can someone suggest a solution?



Also if possible, could someone tell me what are the conditions for when the inequalities are preserved under logarithms and what are the properties that need to be satisfied by the base? EDIT: Got the answer for this! Thank you!










share|cite|improve this question




















  • 4




    It's possible to show that if the function you're using is strictly monotonically increasing (which the logarithm is), it will preserve inequalities. If it's strictly monotonically decreasing, it will reverse the inequality. If it's neither, you will destroy the inequality.
    – Adrian Keister
    Nov 14 at 19:41










  • If $varepsilon>1$, the inequality will be preserved— but if $0<varepsilon<1$, the inequality will be reversed.
    – Mercy King
    Nov 14 at 19:47










  • @AdrianKeister and Mercy King : Thank you for your comments. So does that mean the value of $x$ makes no difference and does not have an impact on how the inequality will be upon taking the logarithm? EDIT : Also, what about $varepsilon=1$? Do you just deal with that as a special case?
    – Bhavesh Singhal
    Nov 14 at 19:59












  • Rather than take $log_{varepsilon},$ I would just do $ln$ of both sides.
    – Adrian Keister
    Nov 14 at 20:12










  • @MercyKing: I think the conditions on the variables present would make that an impossibility, right? The LHS is non-negative, and $k^k>0$ for all $kinmathbb{N}$, so $varepsilon<0$ would make the inequality a contradiction.
    – Adrian Keister
    Nov 14 at 20:14













up vote
1
down vote

favorite









up vote
1
down vote

favorite











I am trying to solved this inequality for $k$.



$x^{2k}<varepsiloncdot k^k$



Here $kinmathbb{N}$ and $x,varepsilon$ are fixed such that $x,varepsiloninmathbb{R}$ and $varepsilon>0$. I was thinking about taking $log_varepsilon$ on both sides but I am not sure about whether this will preserve the inequality or not? Moreover, I cannot really think of any other way to isolate $k$ other than to consider logarithms.



EDIT: Can someone suggest a solution?



Also if possible, could someone tell me what are the conditions for when the inequalities are preserved under logarithms and what are the properties that need to be satisfied by the base? EDIT: Got the answer for this! Thank you!










share|cite|improve this question















I am trying to solved this inequality for $k$.



$x^{2k}<varepsiloncdot k^k$



Here $kinmathbb{N}$ and $x,varepsilon$ are fixed such that $x,varepsiloninmathbb{R}$ and $varepsilon>0$. I was thinking about taking $log_varepsilon$ on both sides but I am not sure about whether this will preserve the inequality or not? Moreover, I cannot really think of any other way to isolate $k$ other than to consider logarithms.



EDIT: Can someone suggest a solution?



Also if possible, could someone tell me what are the conditions for when the inequalities are preserved under logarithms and what are the properties that need to be satisfied by the base? EDIT: Got the answer for this! Thank you!







algebra-precalculus functions






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edited Nov 14 at 20:47

























asked Nov 14 at 19:39









Bhavesh Singhal

184




184








  • 4




    It's possible to show that if the function you're using is strictly monotonically increasing (which the logarithm is), it will preserve inequalities. If it's strictly monotonically decreasing, it will reverse the inequality. If it's neither, you will destroy the inequality.
    – Adrian Keister
    Nov 14 at 19:41










  • If $varepsilon>1$, the inequality will be preserved— but if $0<varepsilon<1$, the inequality will be reversed.
    – Mercy King
    Nov 14 at 19:47










  • @AdrianKeister and Mercy King : Thank you for your comments. So does that mean the value of $x$ makes no difference and does not have an impact on how the inequality will be upon taking the logarithm? EDIT : Also, what about $varepsilon=1$? Do you just deal with that as a special case?
    – Bhavesh Singhal
    Nov 14 at 19:59












  • Rather than take $log_{varepsilon},$ I would just do $ln$ of both sides.
    – Adrian Keister
    Nov 14 at 20:12










  • @MercyKing: I think the conditions on the variables present would make that an impossibility, right? The LHS is non-negative, and $k^k>0$ for all $kinmathbb{N}$, so $varepsilon<0$ would make the inequality a contradiction.
    – Adrian Keister
    Nov 14 at 20:14














  • 4




    It's possible to show that if the function you're using is strictly monotonically increasing (which the logarithm is), it will preserve inequalities. If it's strictly monotonically decreasing, it will reverse the inequality. If it's neither, you will destroy the inequality.
    – Adrian Keister
    Nov 14 at 19:41










  • If $varepsilon>1$, the inequality will be preserved— but if $0<varepsilon<1$, the inequality will be reversed.
    – Mercy King
    Nov 14 at 19:47










  • @AdrianKeister and Mercy King : Thank you for your comments. So does that mean the value of $x$ makes no difference and does not have an impact on how the inequality will be upon taking the logarithm? EDIT : Also, what about $varepsilon=1$? Do you just deal with that as a special case?
    – Bhavesh Singhal
    Nov 14 at 19:59












  • Rather than take $log_{varepsilon},$ I would just do $ln$ of both sides.
    – Adrian Keister
    Nov 14 at 20:12










  • @MercyKing: I think the conditions on the variables present would make that an impossibility, right? The LHS is non-negative, and $k^k>0$ for all $kinmathbb{N}$, so $varepsilon<0$ would make the inequality a contradiction.
    – Adrian Keister
    Nov 14 at 20:14








4




4




It's possible to show that if the function you're using is strictly monotonically increasing (which the logarithm is), it will preserve inequalities. If it's strictly monotonically decreasing, it will reverse the inequality. If it's neither, you will destroy the inequality.
– Adrian Keister
Nov 14 at 19:41




It's possible to show that if the function you're using is strictly monotonically increasing (which the logarithm is), it will preserve inequalities. If it's strictly monotonically decreasing, it will reverse the inequality. If it's neither, you will destroy the inequality.
– Adrian Keister
Nov 14 at 19:41












If $varepsilon>1$, the inequality will be preserved— but if $0<varepsilon<1$, the inequality will be reversed.
– Mercy King
Nov 14 at 19:47




If $varepsilon>1$, the inequality will be preserved— but if $0<varepsilon<1$, the inequality will be reversed.
– Mercy King
Nov 14 at 19:47












@AdrianKeister and Mercy King : Thank you for your comments. So does that mean the value of $x$ makes no difference and does not have an impact on how the inequality will be upon taking the logarithm? EDIT : Also, what about $varepsilon=1$? Do you just deal with that as a special case?
– Bhavesh Singhal
Nov 14 at 19:59






@AdrianKeister and Mercy King : Thank you for your comments. So does that mean the value of $x$ makes no difference and does not have an impact on how the inequality will be upon taking the logarithm? EDIT : Also, what about $varepsilon=1$? Do you just deal with that as a special case?
– Bhavesh Singhal
Nov 14 at 19:59














Rather than take $log_{varepsilon},$ I would just do $ln$ of both sides.
– Adrian Keister
Nov 14 at 20:12




Rather than take $log_{varepsilon},$ I would just do $ln$ of both sides.
– Adrian Keister
Nov 14 at 20:12












@MercyKing: I think the conditions on the variables present would make that an impossibility, right? The LHS is non-negative, and $k^k>0$ for all $kinmathbb{N}$, so $varepsilon<0$ would make the inequality a contradiction.
– Adrian Keister
Nov 14 at 20:14




@MercyKing: I think the conditions on the variables present would make that an impossibility, right? The LHS is non-negative, and $k^k>0$ for all $kinmathbb{N}$, so $varepsilon<0$ would make the inequality a contradiction.
– Adrian Keister
Nov 14 at 20:14










4 Answers
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up vote
1
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Suppose $0 < a < c$ and $b = 1$. The $b^k = 1$ for all $k$ and defining $log_1$ makes no sense. So you can't use logarithms base $1$. (Duh....)



If $b > 1$ then for every $x > 0$ there is some $k = log_b x$ so that $b^k = x$.



Notice that if $0 < x < 1$ then $k =log_b x < 0$. If $x = 1$ then $k = log_b x = 0$. And if $x > 1$ then $k=log_b x > 0$.



Now if we are given that $a,c$ are positive then notice:



$a < c iff 1 < frac ca iff log_b frac ca > 0 iff log_b c - log_b a > 0 iff log_b a < log_b c$.



So if $b > 1$ then you can "take the logs of both sides" to perserve inequality.



If $0 < b < 1$ then we still have for every $x > 0$ there is some $k = log_b x$ so that $b^k = x$.



If this is not as intuitively obvious note if $0 < b < 1$ then $M = frac 1b > 0$ and there is $j = log_M x$ so that $M^j = x$. So $(frac 1b)^{j} = b^{-j} = x$ so $k = log_b x = -j = -log_M x$. So it is true.



We get "double negatives" but...



If $x > 1$ then $b^k = x> 1implies k =log_b x < 0$. Of $x = 1$ then $b^k = x=1implies k = log_b x = 0$ and if $0 < x < 1$ then $0<b^k = x< 1implies k = log_b x > 0$.



And again if $a,c$ are positive and $a < ciff 1 < frac caiff log_b frac ca <0 iff log_b c - log_b a <0 iff log_b a > log_b c$.



So....



Upshot. You can take logs of both sides to preserve inequality if the base is more than $1$. You can take logs of both sides to reverse inequality if the base is less that $1$. And if the base is $1$ you can't do anything.






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    up vote
    0
    down vote













    Because the log function (to a base greater than $1$) is monotonically increasing, you can take the log of both sides. You still won't get a clean answer for $k$ given $x,varepsilon$ If we make it an equality to find the minimum $k$ we can look at it as $$left(frac {x^2}kright)^k=varepsilon$$



    If $varepsilon$ is a small positive number as it usually is, we can see that $k$ must be greater than $x^2$. If $x^2$ is reasonably large, $k$ will not be much larger because the power will be so high. If $x=10, varepsilon=10^{-6}$ we have $k approx 113$






    share|cite|improve this answer





















    • Could you use the Lambert $W$ function here? Looks like you might, but I'm not sure $W$ is monotonic; you might lose the inequality.
      – Adrian Keister
      Nov 14 at 22:29












    • From your equation, Mathematica yields $k=-ln(varepsilon)/W(-ln(varepsilon)/x^2).$
      – Adrian Keister
      Nov 14 at 22:32










    • @AdrianKeister: Mathematica says so, but I haven't understood the W function to say one way or the other.
      – Ross Millikan
      Nov 14 at 22:47


















    up vote
    0
    down vote



    accepted










    Choosing $k=max left( lceil x^2rceil+1, lceilfrac{x^2}{varepsilon}rceil right)$ works. Thanks to everyone for their suggestions.






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      Since $ln$ is increasing, we have
      $$
      2kln(|x|)<kvarepsilonln(k)
      $$

      i.e
      $$
      frac{2kln(|x|)}{kvarepsilon}=2varepsilon^{-1}ln(|x|)<ln(k)
      $$

      Hence
      $$
      k>|x|^{2/varepsilon}
      $$






      share|cite|improve this answer





















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        4 Answers
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        4 Answers
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        Suppose $0 < a < c$ and $b = 1$. The $b^k = 1$ for all $k$ and defining $log_1$ makes no sense. So you can't use logarithms base $1$. (Duh....)



        If $b > 1$ then for every $x > 0$ there is some $k = log_b x$ so that $b^k = x$.



        Notice that if $0 < x < 1$ then $k =log_b x < 0$. If $x = 1$ then $k = log_b x = 0$. And if $x > 1$ then $k=log_b x > 0$.



        Now if we are given that $a,c$ are positive then notice:



        $a < c iff 1 < frac ca iff log_b frac ca > 0 iff log_b c - log_b a > 0 iff log_b a < log_b c$.



        So if $b > 1$ then you can "take the logs of both sides" to perserve inequality.



        If $0 < b < 1$ then we still have for every $x > 0$ there is some $k = log_b x$ so that $b^k = x$.



        If this is not as intuitively obvious note if $0 < b < 1$ then $M = frac 1b > 0$ and there is $j = log_M x$ so that $M^j = x$. So $(frac 1b)^{j} = b^{-j} = x$ so $k = log_b x = -j = -log_M x$. So it is true.



        We get "double negatives" but...



        If $x > 1$ then $b^k = x> 1implies k =log_b x < 0$. Of $x = 1$ then $b^k = x=1implies k = log_b x = 0$ and if $0 < x < 1$ then $0<b^k = x< 1implies k = log_b x > 0$.



        And again if $a,c$ are positive and $a < ciff 1 < frac caiff log_b frac ca <0 iff log_b c - log_b a <0 iff log_b a > log_b c$.



        So....



        Upshot. You can take logs of both sides to preserve inequality if the base is more than $1$. You can take logs of both sides to reverse inequality if the base is less that $1$. And if the base is $1$ you can't do anything.






        share|cite|improve this answer

























          up vote
          1
          down vote













          Suppose $0 < a < c$ and $b = 1$. The $b^k = 1$ for all $k$ and defining $log_1$ makes no sense. So you can't use logarithms base $1$. (Duh....)



          If $b > 1$ then for every $x > 0$ there is some $k = log_b x$ so that $b^k = x$.



          Notice that if $0 < x < 1$ then $k =log_b x < 0$. If $x = 1$ then $k = log_b x = 0$. And if $x > 1$ then $k=log_b x > 0$.



          Now if we are given that $a,c$ are positive then notice:



          $a < c iff 1 < frac ca iff log_b frac ca > 0 iff log_b c - log_b a > 0 iff log_b a < log_b c$.



          So if $b > 1$ then you can "take the logs of both sides" to perserve inequality.



          If $0 < b < 1$ then we still have for every $x > 0$ there is some $k = log_b x$ so that $b^k = x$.



          If this is not as intuitively obvious note if $0 < b < 1$ then $M = frac 1b > 0$ and there is $j = log_M x$ so that $M^j = x$. So $(frac 1b)^{j} = b^{-j} = x$ so $k = log_b x = -j = -log_M x$. So it is true.



          We get "double negatives" but...



          If $x > 1$ then $b^k = x> 1implies k =log_b x < 0$. Of $x = 1$ then $b^k = x=1implies k = log_b x = 0$ and if $0 < x < 1$ then $0<b^k = x< 1implies k = log_b x > 0$.



          And again if $a,c$ are positive and $a < ciff 1 < frac caiff log_b frac ca <0 iff log_b c - log_b a <0 iff log_b a > log_b c$.



          So....



          Upshot. You can take logs of both sides to preserve inequality if the base is more than $1$. You can take logs of both sides to reverse inequality if the base is less that $1$. And if the base is $1$ you can't do anything.






          share|cite|improve this answer























            up vote
            1
            down vote










            up vote
            1
            down vote









            Suppose $0 < a < c$ and $b = 1$. The $b^k = 1$ for all $k$ and defining $log_1$ makes no sense. So you can't use logarithms base $1$. (Duh....)



            If $b > 1$ then for every $x > 0$ there is some $k = log_b x$ so that $b^k = x$.



            Notice that if $0 < x < 1$ then $k =log_b x < 0$. If $x = 1$ then $k = log_b x = 0$. And if $x > 1$ then $k=log_b x > 0$.



            Now if we are given that $a,c$ are positive then notice:



            $a < c iff 1 < frac ca iff log_b frac ca > 0 iff log_b c - log_b a > 0 iff log_b a < log_b c$.



            So if $b > 1$ then you can "take the logs of both sides" to perserve inequality.



            If $0 < b < 1$ then we still have for every $x > 0$ there is some $k = log_b x$ so that $b^k = x$.



            If this is not as intuitively obvious note if $0 < b < 1$ then $M = frac 1b > 0$ and there is $j = log_M x$ so that $M^j = x$. So $(frac 1b)^{j} = b^{-j} = x$ so $k = log_b x = -j = -log_M x$. So it is true.



            We get "double negatives" but...



            If $x > 1$ then $b^k = x> 1implies k =log_b x < 0$. Of $x = 1$ then $b^k = x=1implies k = log_b x = 0$ and if $0 < x < 1$ then $0<b^k = x< 1implies k = log_b x > 0$.



            And again if $a,c$ are positive and $a < ciff 1 < frac caiff log_b frac ca <0 iff log_b c - log_b a <0 iff log_b a > log_b c$.



            So....



            Upshot. You can take logs of both sides to preserve inequality if the base is more than $1$. You can take logs of both sides to reverse inequality if the base is less that $1$. And if the base is $1$ you can't do anything.






            share|cite|improve this answer












            Suppose $0 < a < c$ and $b = 1$. The $b^k = 1$ for all $k$ and defining $log_1$ makes no sense. So you can't use logarithms base $1$. (Duh....)



            If $b > 1$ then for every $x > 0$ there is some $k = log_b x$ so that $b^k = x$.



            Notice that if $0 < x < 1$ then $k =log_b x < 0$. If $x = 1$ then $k = log_b x = 0$. And if $x > 1$ then $k=log_b x > 0$.



            Now if we are given that $a,c$ are positive then notice:



            $a < c iff 1 < frac ca iff log_b frac ca > 0 iff log_b c - log_b a > 0 iff log_b a < log_b c$.



            So if $b > 1$ then you can "take the logs of both sides" to perserve inequality.



            If $0 < b < 1$ then we still have for every $x > 0$ there is some $k = log_b x$ so that $b^k = x$.



            If this is not as intuitively obvious note if $0 < b < 1$ then $M = frac 1b > 0$ and there is $j = log_M x$ so that $M^j = x$. So $(frac 1b)^{j} = b^{-j} = x$ so $k = log_b x = -j = -log_M x$. So it is true.



            We get "double negatives" but...



            If $x > 1$ then $b^k = x> 1implies k =log_b x < 0$. Of $x = 1$ then $b^k = x=1implies k = log_b x = 0$ and if $0 < x < 1$ then $0<b^k = x< 1implies k = log_b x > 0$.



            And again if $a,c$ are positive and $a < ciff 1 < frac caiff log_b frac ca <0 iff log_b c - log_b a <0 iff log_b a > log_b c$.



            So....



            Upshot. You can take logs of both sides to preserve inequality if the base is more than $1$. You can take logs of both sides to reverse inequality if the base is less that $1$. And if the base is $1$ you can't do anything.







            share|cite|improve this answer












            share|cite|improve this answer



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            answered Nov 15 at 19:38









            fleablood

            65.7k22682




            65.7k22682






















                up vote
                0
                down vote













                Because the log function (to a base greater than $1$) is monotonically increasing, you can take the log of both sides. You still won't get a clean answer for $k$ given $x,varepsilon$ If we make it an equality to find the minimum $k$ we can look at it as $$left(frac {x^2}kright)^k=varepsilon$$



                If $varepsilon$ is a small positive number as it usually is, we can see that $k$ must be greater than $x^2$. If $x^2$ is reasonably large, $k$ will not be much larger because the power will be so high. If $x=10, varepsilon=10^{-6}$ we have $k approx 113$






                share|cite|improve this answer





















                • Could you use the Lambert $W$ function here? Looks like you might, but I'm not sure $W$ is monotonic; you might lose the inequality.
                  – Adrian Keister
                  Nov 14 at 22:29












                • From your equation, Mathematica yields $k=-ln(varepsilon)/W(-ln(varepsilon)/x^2).$
                  – Adrian Keister
                  Nov 14 at 22:32










                • @AdrianKeister: Mathematica says so, but I haven't understood the W function to say one way or the other.
                  – Ross Millikan
                  Nov 14 at 22:47















                up vote
                0
                down vote













                Because the log function (to a base greater than $1$) is monotonically increasing, you can take the log of both sides. You still won't get a clean answer for $k$ given $x,varepsilon$ If we make it an equality to find the minimum $k$ we can look at it as $$left(frac {x^2}kright)^k=varepsilon$$



                If $varepsilon$ is a small positive number as it usually is, we can see that $k$ must be greater than $x^2$. If $x^2$ is reasonably large, $k$ will not be much larger because the power will be so high. If $x=10, varepsilon=10^{-6}$ we have $k approx 113$






                share|cite|improve this answer





















                • Could you use the Lambert $W$ function here? Looks like you might, but I'm not sure $W$ is monotonic; you might lose the inequality.
                  – Adrian Keister
                  Nov 14 at 22:29












                • From your equation, Mathematica yields $k=-ln(varepsilon)/W(-ln(varepsilon)/x^2).$
                  – Adrian Keister
                  Nov 14 at 22:32










                • @AdrianKeister: Mathematica says so, but I haven't understood the W function to say one way or the other.
                  – Ross Millikan
                  Nov 14 at 22:47













                up vote
                0
                down vote










                up vote
                0
                down vote









                Because the log function (to a base greater than $1$) is monotonically increasing, you can take the log of both sides. You still won't get a clean answer for $k$ given $x,varepsilon$ If we make it an equality to find the minimum $k$ we can look at it as $$left(frac {x^2}kright)^k=varepsilon$$



                If $varepsilon$ is a small positive number as it usually is, we can see that $k$ must be greater than $x^2$. If $x^2$ is reasonably large, $k$ will not be much larger because the power will be so high. If $x=10, varepsilon=10^{-6}$ we have $k approx 113$






                share|cite|improve this answer












                Because the log function (to a base greater than $1$) is monotonically increasing, you can take the log of both sides. You still won't get a clean answer for $k$ given $x,varepsilon$ If we make it an equality to find the minimum $k$ we can look at it as $$left(frac {x^2}kright)^k=varepsilon$$



                If $varepsilon$ is a small positive number as it usually is, we can see that $k$ must be greater than $x^2$. If $x^2$ is reasonably large, $k$ will not be much larger because the power will be so high. If $x=10, varepsilon=10^{-6}$ we have $k approx 113$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 14 at 21:29









                Ross Millikan

                287k23195364




                287k23195364












                • Could you use the Lambert $W$ function here? Looks like you might, but I'm not sure $W$ is monotonic; you might lose the inequality.
                  – Adrian Keister
                  Nov 14 at 22:29












                • From your equation, Mathematica yields $k=-ln(varepsilon)/W(-ln(varepsilon)/x^2).$
                  – Adrian Keister
                  Nov 14 at 22:32










                • @AdrianKeister: Mathematica says so, but I haven't understood the W function to say one way or the other.
                  – Ross Millikan
                  Nov 14 at 22:47


















                • Could you use the Lambert $W$ function here? Looks like you might, but I'm not sure $W$ is monotonic; you might lose the inequality.
                  – Adrian Keister
                  Nov 14 at 22:29












                • From your equation, Mathematica yields $k=-ln(varepsilon)/W(-ln(varepsilon)/x^2).$
                  – Adrian Keister
                  Nov 14 at 22:32










                • @AdrianKeister: Mathematica says so, but I haven't understood the W function to say one way or the other.
                  – Ross Millikan
                  Nov 14 at 22:47
















                Could you use the Lambert $W$ function here? Looks like you might, but I'm not sure $W$ is monotonic; you might lose the inequality.
                – Adrian Keister
                Nov 14 at 22:29






                Could you use the Lambert $W$ function here? Looks like you might, but I'm not sure $W$ is monotonic; you might lose the inequality.
                – Adrian Keister
                Nov 14 at 22:29














                From your equation, Mathematica yields $k=-ln(varepsilon)/W(-ln(varepsilon)/x^2).$
                – Adrian Keister
                Nov 14 at 22:32




                From your equation, Mathematica yields $k=-ln(varepsilon)/W(-ln(varepsilon)/x^2).$
                – Adrian Keister
                Nov 14 at 22:32












                @AdrianKeister: Mathematica says so, but I haven't understood the W function to say one way or the other.
                – Ross Millikan
                Nov 14 at 22:47




                @AdrianKeister: Mathematica says so, but I haven't understood the W function to say one way or the other.
                – Ross Millikan
                Nov 14 at 22:47










                up vote
                0
                down vote



                accepted










                Choosing $k=max left( lceil x^2rceil+1, lceilfrac{x^2}{varepsilon}rceil right)$ works. Thanks to everyone for their suggestions.






                share|cite|improve this answer

























                  up vote
                  0
                  down vote



                  accepted










                  Choosing $k=max left( lceil x^2rceil+1, lceilfrac{x^2}{varepsilon}rceil right)$ works. Thanks to everyone for their suggestions.






                  share|cite|improve this answer























                    up vote
                    0
                    down vote



                    accepted







                    up vote
                    0
                    down vote



                    accepted






                    Choosing $k=max left( lceil x^2rceil+1, lceilfrac{x^2}{varepsilon}rceil right)$ works. Thanks to everyone for their suggestions.






                    share|cite|improve this answer












                    Choosing $k=max left( lceil x^2rceil+1, lceilfrac{x^2}{varepsilon}rceil right)$ works. Thanks to everyone for their suggestions.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 15 at 18:32









                    Bhavesh Singhal

                    184




                    184






















                        up vote
                        0
                        down vote













                        Since $ln$ is increasing, we have
                        $$
                        2kln(|x|)<kvarepsilonln(k)
                        $$

                        i.e
                        $$
                        frac{2kln(|x|)}{kvarepsilon}=2varepsilon^{-1}ln(|x|)<ln(k)
                        $$

                        Hence
                        $$
                        k>|x|^{2/varepsilon}
                        $$






                        share|cite|improve this answer

























                          up vote
                          0
                          down vote













                          Since $ln$ is increasing, we have
                          $$
                          2kln(|x|)<kvarepsilonln(k)
                          $$

                          i.e
                          $$
                          frac{2kln(|x|)}{kvarepsilon}=2varepsilon^{-1}ln(|x|)<ln(k)
                          $$

                          Hence
                          $$
                          k>|x|^{2/varepsilon}
                          $$






                          share|cite|improve this answer























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            Since $ln$ is increasing, we have
                            $$
                            2kln(|x|)<kvarepsilonln(k)
                            $$

                            i.e
                            $$
                            frac{2kln(|x|)}{kvarepsilon}=2varepsilon^{-1}ln(|x|)<ln(k)
                            $$

                            Hence
                            $$
                            k>|x|^{2/varepsilon}
                            $$






                            share|cite|improve this answer












                            Since $ln$ is increasing, we have
                            $$
                            2kln(|x|)<kvarepsilonln(k)
                            $$

                            i.e
                            $$
                            frac{2kln(|x|)}{kvarepsilon}=2varepsilon^{-1}ln(|x|)<ln(k)
                            $$

                            Hence
                            $$
                            k>|x|^{2/varepsilon}
                            $$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 16 at 0:23









                            Mercy King

                            13.9k11327




                            13.9k11327






























                                 

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