How do I describe a closed form of the transition from ln(x) to 1/x?












1














I am trying to write a closed form for the $n$th derivative of $x log(x)$.



$frac{d}{dx} x log(x)= log(x)+1$



$frac{d^{2}}{dx^{2}}= frac{1}{x}$



and then from here there are no $log(x)$ terms ever again, it might as well be the $m$th derivative of $frac{1}{x}$.
So...do I need the ever-daunted step function? Or how can I describe this nth derivative in closed form?










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  • Why would you need to use the step function for $frac 1 x$?
    – Sudix
    1 hour ago










  • Why wouldn't I? $log(x)$ exists for the 0th and 1st derivative, then suddenly becomes 0. If you can recommend something better you're free to do that.
    – user14554
    1 hour ago












  • Just derive $frac 1 x$, as you'd do if there never was a $log$-function to begin with?
    – Sudix
    1 hour ago










  • Do you just ignore the premise of a statement for everything else in math? Just the $frac{1}{x}$ series has already been derived and you can google it easily, but I have not seen anything that generalizes the transition from $log$ to no $log$.
    – user14554
    34 mins ago












  • I'm not seeing this premise, that's why I'm asking. What is it?
    – Sudix
    32 mins ago
















1














I am trying to write a closed form for the $n$th derivative of $x log(x)$.



$frac{d}{dx} x log(x)= log(x)+1$



$frac{d^{2}}{dx^{2}}= frac{1}{x}$



and then from here there are no $log(x)$ terms ever again, it might as well be the $m$th derivative of $frac{1}{x}$.
So...do I need the ever-daunted step function? Or how can I describe this nth derivative in closed form?










share|cite|improve this question







New contributor




user14554 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • Why would you need to use the step function for $frac 1 x$?
    – Sudix
    1 hour ago










  • Why wouldn't I? $log(x)$ exists for the 0th and 1st derivative, then suddenly becomes 0. If you can recommend something better you're free to do that.
    – user14554
    1 hour ago












  • Just derive $frac 1 x$, as you'd do if there never was a $log$-function to begin with?
    – Sudix
    1 hour ago










  • Do you just ignore the premise of a statement for everything else in math? Just the $frac{1}{x}$ series has already been derived and you can google it easily, but I have not seen anything that generalizes the transition from $log$ to no $log$.
    – user14554
    34 mins ago












  • I'm not seeing this premise, that's why I'm asking. What is it?
    – Sudix
    32 mins ago














1












1








1


1





I am trying to write a closed form for the $n$th derivative of $x log(x)$.



$frac{d}{dx} x log(x)= log(x)+1$



$frac{d^{2}}{dx^{2}}= frac{1}{x}$



and then from here there are no $log(x)$ terms ever again, it might as well be the $m$th derivative of $frac{1}{x}$.
So...do I need the ever-daunted step function? Or how can I describe this nth derivative in closed form?










share|cite|improve this question







New contributor




user14554 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I am trying to write a closed form for the $n$th derivative of $x log(x)$.



$frac{d}{dx} x log(x)= log(x)+1$



$frac{d^{2}}{dx^{2}}= frac{1}{x}$



and then from here there are no $log(x)$ terms ever again, it might as well be the $m$th derivative of $frac{1}{x}$.
So...do I need the ever-daunted step function? Or how can I describe this nth derivative in closed form?







derivatives






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New contributor




user14554 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




user14554 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






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Check out our Code of Conduct.









asked 1 hour ago









user14554

91




91




New contributor




user14554 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





user14554 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






user14554 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • Why would you need to use the step function for $frac 1 x$?
    – Sudix
    1 hour ago










  • Why wouldn't I? $log(x)$ exists for the 0th and 1st derivative, then suddenly becomes 0. If you can recommend something better you're free to do that.
    – user14554
    1 hour ago












  • Just derive $frac 1 x$, as you'd do if there never was a $log$-function to begin with?
    – Sudix
    1 hour ago










  • Do you just ignore the premise of a statement for everything else in math? Just the $frac{1}{x}$ series has already been derived and you can google it easily, but I have not seen anything that generalizes the transition from $log$ to no $log$.
    – user14554
    34 mins ago












  • I'm not seeing this premise, that's why I'm asking. What is it?
    – Sudix
    32 mins ago


















  • Why would you need to use the step function for $frac 1 x$?
    – Sudix
    1 hour ago










  • Why wouldn't I? $log(x)$ exists for the 0th and 1st derivative, then suddenly becomes 0. If you can recommend something better you're free to do that.
    – user14554
    1 hour ago












  • Just derive $frac 1 x$, as you'd do if there never was a $log$-function to begin with?
    – Sudix
    1 hour ago










  • Do you just ignore the premise of a statement for everything else in math? Just the $frac{1}{x}$ series has already been derived and you can google it easily, but I have not seen anything that generalizes the transition from $log$ to no $log$.
    – user14554
    34 mins ago












  • I'm not seeing this premise, that's why I'm asking. What is it?
    – Sudix
    32 mins ago
















Why would you need to use the step function for $frac 1 x$?
– Sudix
1 hour ago




Why would you need to use the step function for $frac 1 x$?
– Sudix
1 hour ago












Why wouldn't I? $log(x)$ exists for the 0th and 1st derivative, then suddenly becomes 0. If you can recommend something better you're free to do that.
– user14554
1 hour ago






Why wouldn't I? $log(x)$ exists for the 0th and 1st derivative, then suddenly becomes 0. If you can recommend something better you're free to do that.
– user14554
1 hour ago














Just derive $frac 1 x$, as you'd do if there never was a $log$-function to begin with?
– Sudix
1 hour ago




Just derive $frac 1 x$, as you'd do if there never was a $log$-function to begin with?
– Sudix
1 hour ago












Do you just ignore the premise of a statement for everything else in math? Just the $frac{1}{x}$ series has already been derived and you can google it easily, but I have not seen anything that generalizes the transition from $log$ to no $log$.
– user14554
34 mins ago






Do you just ignore the premise of a statement for everything else in math? Just the $frac{1}{x}$ series has already been derived and you can google it easily, but I have not seen anything that generalizes the transition from $log$ to no $log$.
– user14554
34 mins ago














I'm not seeing this premise, that's why I'm asking. What is it?
– Sudix
32 mins ago




I'm not seeing this premise, that's why I'm asking. What is it?
– Sudix
32 mins ago










3 Answers
3






active

oldest

votes


















2














$$ y'=ln x+1$$



And for $n>1$,



$$ y'' = 1/x =x^{-1}$$



$$y'''=-x^{-2}$$
$$y^{(4)} = 2 x^{-3}$$$$ y^{(n)} = (-1)^{n}(n-2)!x^{-n+1}$$






share|cite|improve this answer























  • You're missing a factor there
    – Sudix
    50 mins ago










  • Thanks for the comment. I fixed it.
    – Mohammad Riazi-Kermani
    42 mins ago










  • So I plugged in n=1 to your formula and it doesn't exist because (-1)! doesn't exist. The whole point of this thread is to capture a single closed form for all n, no loophole n>1 terms, those are trivial.
    – user14554
    38 mins ago





















1














If $n$ is a positive integer, the function $h(n)=lfloor frac{1}{n} rfloor$ takes values $1$ for $n=1$ and $0$ for $n>1$. Therefore, using computations from @Mohammad Riazi-Kermani's answer, we have



$$
frac{d^n}{dx^n}(xln(x)) = h(n)(1+ln(x))+(1-h(n))(-1)^n(n-2)!x^{-n+1}
$$



for any $ngeqslant 1$.



IMHO, it is slightly annoying to have to deal with the special case $n=1$ but I don't think that the closed formula is more convenient. One may also argue that using the floor function is somehow cheating, since the floor function is just a (very convenient) way to disguise a piecewise function...






share|cite|improve this answer





















  • It's better than what I thought of, the floor function isn't unconventional in dealing with integers.
    – user14554
    18 mins ago










  • Yeah when such a weird piece-wise-behaving function is required, I've never seen something that doesn't use the floor function or something like that; I'm starting to think its impossible.
    – Ovi
    16 mins ago





















1














So basically, you are asking for a formula that is not piecewise-defined, right?



If so, one good way to do this is to use power series, since they are very easy to differentiate to arbitrary orders, even if the original function has a complicated mix of transcendental functions, as the terms are usually simply polynomial. To maximize the easiness, however, we might first want to make a substitution to "conjugate" the differentiation problem to a simpler one. Namely, we can use the Mercator series for $log(1 + x)$, which gives



$$log(1 + x) = sum_{n=1}^{infty} frac{(-1)^{n-1}}{n} x^n$$



We first rearrange this a bit, setting $x mapsto x - 1$, so that



$$log(x) = sum_{n=1}^{infty} frac{(-1)^{n-1}}{n} (x - 1)^n$$



thus



$$x log(x) = sum_{n=1}^{infty} frac{(-1)^{n-1}}{n} x(x - 1)^n$$



and so the problem now reduces to differentiating the polynomial expression $x(x - 1)^n$, i.e. finding



$$frac{d^m}{dx^m} [x(x - 1)^n]$$



in terms of $m$. We can expand this using the binomial theorem to get



$$x(x - 1)^n = sum_{k=0}^{n} binom{n}{k} (-1)^{n-k} x^{k+1}$$



and now we can differentiate $x^k$ $m$ times using the formula



$$frac{d^m}{dx^m} x^k = (k)_m x^{k-1}$$



where $(k)_m$ is the "falling factorial" given by $(a)_n = a(a-1)(a-2)cdots(a-n+1)$, which has $n$ factors. Note that if $m > k$, this is zero, and the derivatives thus go to zero.



Thus



$$frac{d^m}{dx^m} x(x - 1)^n = sum_{k=0}^{n} binom{n}{k} (-1)^{n-k} (k+1)_m x^{k+1-m}$$



and the original is



$$frac{d^m}{dx^m} [x log(x)] = sum_{n=1}^{infty} frac{(-1)^{n-1}}{n} left[sum_{k=0}^{n} binom{n}{k} (-1)^{n-k} (k+1)_m x^{k+1-m}right]$$



for which a rearrangement gives



$$frac{d^m}{dx^m} [x log(x)] = sum_{n=1}^{infty} sum_{k=0}^{n} binom{n}{k} frac{(-1)^{k+1} (k+1)_m}{n} x^{k+1-m}$$



This series will, of course, only converge in $x in (0, 2)$, I believe, but this is the closest thing to a "closed form" you will get even though it's not technically one, but moreover you can even vary $m$ smoothly, if you use the extension $(k+1)_m = frac{Gamma(k+2)}{Gamma(k-m-2)}$ and this will vary smoothly as well, so that you will have a continuous transition from $ln x$ to $frac{1}{x}$, and thus describes, or better, interpolates, the "transition" you ask about, and uses no ad-hoc step functions.



There is probably also some way to express the series using some kind of hypergeometric functions, but I don't feel like trying to rearrange it right now.






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    3 Answers
    3






    active

    oldest

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    3 Answers
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    active

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    active

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    oldest

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    2














    $$ y'=ln x+1$$



    And for $n>1$,



    $$ y'' = 1/x =x^{-1}$$



    $$y'''=-x^{-2}$$
    $$y^{(4)} = 2 x^{-3}$$$$ y^{(n)} = (-1)^{n}(n-2)!x^{-n+1}$$






    share|cite|improve this answer























    • You're missing a factor there
      – Sudix
      50 mins ago










    • Thanks for the comment. I fixed it.
      – Mohammad Riazi-Kermani
      42 mins ago










    • So I plugged in n=1 to your formula and it doesn't exist because (-1)! doesn't exist. The whole point of this thread is to capture a single closed form for all n, no loophole n>1 terms, those are trivial.
      – user14554
      38 mins ago


















    2














    $$ y'=ln x+1$$



    And for $n>1$,



    $$ y'' = 1/x =x^{-1}$$



    $$y'''=-x^{-2}$$
    $$y^{(4)} = 2 x^{-3}$$$$ y^{(n)} = (-1)^{n}(n-2)!x^{-n+1}$$






    share|cite|improve this answer























    • You're missing a factor there
      – Sudix
      50 mins ago










    • Thanks for the comment. I fixed it.
      – Mohammad Riazi-Kermani
      42 mins ago










    • So I plugged in n=1 to your formula and it doesn't exist because (-1)! doesn't exist. The whole point of this thread is to capture a single closed form for all n, no loophole n>1 terms, those are trivial.
      – user14554
      38 mins ago
















    2












    2








    2






    $$ y'=ln x+1$$



    And for $n>1$,



    $$ y'' = 1/x =x^{-1}$$



    $$y'''=-x^{-2}$$
    $$y^{(4)} = 2 x^{-3}$$$$ y^{(n)} = (-1)^{n}(n-2)!x^{-n+1}$$






    share|cite|improve this answer














    $$ y'=ln x+1$$



    And for $n>1$,



    $$ y'' = 1/x =x^{-1}$$



    $$y'''=-x^{-2}$$
    $$y^{(4)} = 2 x^{-3}$$$$ y^{(n)} = (-1)^{n}(n-2)!x^{-n+1}$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 43 mins ago

























    answered 1 hour ago









    Mohammad Riazi-Kermani

    40.5k42058




    40.5k42058












    • You're missing a factor there
      – Sudix
      50 mins ago










    • Thanks for the comment. I fixed it.
      – Mohammad Riazi-Kermani
      42 mins ago










    • So I plugged in n=1 to your formula and it doesn't exist because (-1)! doesn't exist. The whole point of this thread is to capture a single closed form for all n, no loophole n>1 terms, those are trivial.
      – user14554
      38 mins ago




















    • You're missing a factor there
      – Sudix
      50 mins ago










    • Thanks for the comment. I fixed it.
      – Mohammad Riazi-Kermani
      42 mins ago










    • So I plugged in n=1 to your formula and it doesn't exist because (-1)! doesn't exist. The whole point of this thread is to capture a single closed form for all n, no loophole n>1 terms, those are trivial.
      – user14554
      38 mins ago


















    You're missing a factor there
    – Sudix
    50 mins ago




    You're missing a factor there
    – Sudix
    50 mins ago












    Thanks for the comment. I fixed it.
    – Mohammad Riazi-Kermani
    42 mins ago




    Thanks for the comment. I fixed it.
    – Mohammad Riazi-Kermani
    42 mins ago












    So I plugged in n=1 to your formula and it doesn't exist because (-1)! doesn't exist. The whole point of this thread is to capture a single closed form for all n, no loophole n>1 terms, those are trivial.
    – user14554
    38 mins ago






    So I plugged in n=1 to your formula and it doesn't exist because (-1)! doesn't exist. The whole point of this thread is to capture a single closed form for all n, no loophole n>1 terms, those are trivial.
    – user14554
    38 mins ago













    1














    If $n$ is a positive integer, the function $h(n)=lfloor frac{1}{n} rfloor$ takes values $1$ for $n=1$ and $0$ for $n>1$. Therefore, using computations from @Mohammad Riazi-Kermani's answer, we have



    $$
    frac{d^n}{dx^n}(xln(x)) = h(n)(1+ln(x))+(1-h(n))(-1)^n(n-2)!x^{-n+1}
    $$



    for any $ngeqslant 1$.



    IMHO, it is slightly annoying to have to deal with the special case $n=1$ but I don't think that the closed formula is more convenient. One may also argue that using the floor function is somehow cheating, since the floor function is just a (very convenient) way to disguise a piecewise function...






    share|cite|improve this answer





















    • It's better than what I thought of, the floor function isn't unconventional in dealing with integers.
      – user14554
      18 mins ago










    • Yeah when such a weird piece-wise-behaving function is required, I've never seen something that doesn't use the floor function or something like that; I'm starting to think its impossible.
      – Ovi
      16 mins ago


















    1














    If $n$ is a positive integer, the function $h(n)=lfloor frac{1}{n} rfloor$ takes values $1$ for $n=1$ and $0$ for $n>1$. Therefore, using computations from @Mohammad Riazi-Kermani's answer, we have



    $$
    frac{d^n}{dx^n}(xln(x)) = h(n)(1+ln(x))+(1-h(n))(-1)^n(n-2)!x^{-n+1}
    $$



    for any $ngeqslant 1$.



    IMHO, it is slightly annoying to have to deal with the special case $n=1$ but I don't think that the closed formula is more convenient. One may also argue that using the floor function is somehow cheating, since the floor function is just a (very convenient) way to disguise a piecewise function...






    share|cite|improve this answer





















    • It's better than what I thought of, the floor function isn't unconventional in dealing with integers.
      – user14554
      18 mins ago










    • Yeah when such a weird piece-wise-behaving function is required, I've never seen something that doesn't use the floor function or something like that; I'm starting to think its impossible.
      – Ovi
      16 mins ago
















    1












    1








    1






    If $n$ is a positive integer, the function $h(n)=lfloor frac{1}{n} rfloor$ takes values $1$ for $n=1$ and $0$ for $n>1$. Therefore, using computations from @Mohammad Riazi-Kermani's answer, we have



    $$
    frac{d^n}{dx^n}(xln(x)) = h(n)(1+ln(x))+(1-h(n))(-1)^n(n-2)!x^{-n+1}
    $$



    for any $ngeqslant 1$.



    IMHO, it is slightly annoying to have to deal with the special case $n=1$ but I don't think that the closed formula is more convenient. One may also argue that using the floor function is somehow cheating, since the floor function is just a (very convenient) way to disguise a piecewise function...






    share|cite|improve this answer












    If $n$ is a positive integer, the function $h(n)=lfloor frac{1}{n} rfloor$ takes values $1$ for $n=1$ and $0$ for $n>1$. Therefore, using computations from @Mohammad Riazi-Kermani's answer, we have



    $$
    frac{d^n}{dx^n}(xln(x)) = h(n)(1+ln(x))+(1-h(n))(-1)^n(n-2)!x^{-n+1}
    $$



    for any $ngeqslant 1$.



    IMHO, it is slightly annoying to have to deal with the special case $n=1$ but I don't think that the closed formula is more convenient. One may also argue that using the floor function is somehow cheating, since the floor function is just a (very convenient) way to disguise a piecewise function...







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 25 mins ago









    Taladris

    4,65431932




    4,65431932












    • It's better than what I thought of, the floor function isn't unconventional in dealing with integers.
      – user14554
      18 mins ago










    • Yeah when such a weird piece-wise-behaving function is required, I've never seen something that doesn't use the floor function or something like that; I'm starting to think its impossible.
      – Ovi
      16 mins ago




















    • It's better than what I thought of, the floor function isn't unconventional in dealing with integers.
      – user14554
      18 mins ago










    • Yeah when such a weird piece-wise-behaving function is required, I've never seen something that doesn't use the floor function or something like that; I'm starting to think its impossible.
      – Ovi
      16 mins ago


















    It's better than what I thought of, the floor function isn't unconventional in dealing with integers.
    – user14554
    18 mins ago




    It's better than what I thought of, the floor function isn't unconventional in dealing with integers.
    – user14554
    18 mins ago












    Yeah when such a weird piece-wise-behaving function is required, I've never seen something that doesn't use the floor function or something like that; I'm starting to think its impossible.
    – Ovi
    16 mins ago






    Yeah when such a weird piece-wise-behaving function is required, I've never seen something that doesn't use the floor function or something like that; I'm starting to think its impossible.
    – Ovi
    16 mins ago













    1














    So basically, you are asking for a formula that is not piecewise-defined, right?



    If so, one good way to do this is to use power series, since they are very easy to differentiate to arbitrary orders, even if the original function has a complicated mix of transcendental functions, as the terms are usually simply polynomial. To maximize the easiness, however, we might first want to make a substitution to "conjugate" the differentiation problem to a simpler one. Namely, we can use the Mercator series for $log(1 + x)$, which gives



    $$log(1 + x) = sum_{n=1}^{infty} frac{(-1)^{n-1}}{n} x^n$$



    We first rearrange this a bit, setting $x mapsto x - 1$, so that



    $$log(x) = sum_{n=1}^{infty} frac{(-1)^{n-1}}{n} (x - 1)^n$$



    thus



    $$x log(x) = sum_{n=1}^{infty} frac{(-1)^{n-1}}{n} x(x - 1)^n$$



    and so the problem now reduces to differentiating the polynomial expression $x(x - 1)^n$, i.e. finding



    $$frac{d^m}{dx^m} [x(x - 1)^n]$$



    in terms of $m$. We can expand this using the binomial theorem to get



    $$x(x - 1)^n = sum_{k=0}^{n} binom{n}{k} (-1)^{n-k} x^{k+1}$$



    and now we can differentiate $x^k$ $m$ times using the formula



    $$frac{d^m}{dx^m} x^k = (k)_m x^{k-1}$$



    where $(k)_m$ is the "falling factorial" given by $(a)_n = a(a-1)(a-2)cdots(a-n+1)$, which has $n$ factors. Note that if $m > k$, this is zero, and the derivatives thus go to zero.



    Thus



    $$frac{d^m}{dx^m} x(x - 1)^n = sum_{k=0}^{n} binom{n}{k} (-1)^{n-k} (k+1)_m x^{k+1-m}$$



    and the original is



    $$frac{d^m}{dx^m} [x log(x)] = sum_{n=1}^{infty} frac{(-1)^{n-1}}{n} left[sum_{k=0}^{n} binom{n}{k} (-1)^{n-k} (k+1)_m x^{k+1-m}right]$$



    for which a rearrangement gives



    $$frac{d^m}{dx^m} [x log(x)] = sum_{n=1}^{infty} sum_{k=0}^{n} binom{n}{k} frac{(-1)^{k+1} (k+1)_m}{n} x^{k+1-m}$$



    This series will, of course, only converge in $x in (0, 2)$, I believe, but this is the closest thing to a "closed form" you will get even though it's not technically one, but moreover you can even vary $m$ smoothly, if you use the extension $(k+1)_m = frac{Gamma(k+2)}{Gamma(k-m-2)}$ and this will vary smoothly as well, so that you will have a continuous transition from $ln x$ to $frac{1}{x}$, and thus describes, or better, interpolates, the "transition" you ask about, and uses no ad-hoc step functions.



    There is probably also some way to express the series using some kind of hypergeometric functions, but I don't feel like trying to rearrange it right now.






    share|cite|improve this answer


























      1














      So basically, you are asking for a formula that is not piecewise-defined, right?



      If so, one good way to do this is to use power series, since they are very easy to differentiate to arbitrary orders, even if the original function has a complicated mix of transcendental functions, as the terms are usually simply polynomial. To maximize the easiness, however, we might first want to make a substitution to "conjugate" the differentiation problem to a simpler one. Namely, we can use the Mercator series for $log(1 + x)$, which gives



      $$log(1 + x) = sum_{n=1}^{infty} frac{(-1)^{n-1}}{n} x^n$$



      We first rearrange this a bit, setting $x mapsto x - 1$, so that



      $$log(x) = sum_{n=1}^{infty} frac{(-1)^{n-1}}{n} (x - 1)^n$$



      thus



      $$x log(x) = sum_{n=1}^{infty} frac{(-1)^{n-1}}{n} x(x - 1)^n$$



      and so the problem now reduces to differentiating the polynomial expression $x(x - 1)^n$, i.e. finding



      $$frac{d^m}{dx^m} [x(x - 1)^n]$$



      in terms of $m$. We can expand this using the binomial theorem to get



      $$x(x - 1)^n = sum_{k=0}^{n} binom{n}{k} (-1)^{n-k} x^{k+1}$$



      and now we can differentiate $x^k$ $m$ times using the formula



      $$frac{d^m}{dx^m} x^k = (k)_m x^{k-1}$$



      where $(k)_m$ is the "falling factorial" given by $(a)_n = a(a-1)(a-2)cdots(a-n+1)$, which has $n$ factors. Note that if $m > k$, this is zero, and the derivatives thus go to zero.



      Thus



      $$frac{d^m}{dx^m} x(x - 1)^n = sum_{k=0}^{n} binom{n}{k} (-1)^{n-k} (k+1)_m x^{k+1-m}$$



      and the original is



      $$frac{d^m}{dx^m} [x log(x)] = sum_{n=1}^{infty} frac{(-1)^{n-1}}{n} left[sum_{k=0}^{n} binom{n}{k} (-1)^{n-k} (k+1)_m x^{k+1-m}right]$$



      for which a rearrangement gives



      $$frac{d^m}{dx^m} [x log(x)] = sum_{n=1}^{infty} sum_{k=0}^{n} binom{n}{k} frac{(-1)^{k+1} (k+1)_m}{n} x^{k+1-m}$$



      This series will, of course, only converge in $x in (0, 2)$, I believe, but this is the closest thing to a "closed form" you will get even though it's not technically one, but moreover you can even vary $m$ smoothly, if you use the extension $(k+1)_m = frac{Gamma(k+2)}{Gamma(k-m-2)}$ and this will vary smoothly as well, so that you will have a continuous transition from $ln x$ to $frac{1}{x}$, and thus describes, or better, interpolates, the "transition" you ask about, and uses no ad-hoc step functions.



      There is probably also some way to express the series using some kind of hypergeometric functions, but I don't feel like trying to rearrange it right now.






      share|cite|improve this answer
























        1












        1








        1






        So basically, you are asking for a formula that is not piecewise-defined, right?



        If so, one good way to do this is to use power series, since they are very easy to differentiate to arbitrary orders, even if the original function has a complicated mix of transcendental functions, as the terms are usually simply polynomial. To maximize the easiness, however, we might first want to make a substitution to "conjugate" the differentiation problem to a simpler one. Namely, we can use the Mercator series for $log(1 + x)$, which gives



        $$log(1 + x) = sum_{n=1}^{infty} frac{(-1)^{n-1}}{n} x^n$$



        We first rearrange this a bit, setting $x mapsto x - 1$, so that



        $$log(x) = sum_{n=1}^{infty} frac{(-1)^{n-1}}{n} (x - 1)^n$$



        thus



        $$x log(x) = sum_{n=1}^{infty} frac{(-1)^{n-1}}{n} x(x - 1)^n$$



        and so the problem now reduces to differentiating the polynomial expression $x(x - 1)^n$, i.e. finding



        $$frac{d^m}{dx^m} [x(x - 1)^n]$$



        in terms of $m$. We can expand this using the binomial theorem to get



        $$x(x - 1)^n = sum_{k=0}^{n} binom{n}{k} (-1)^{n-k} x^{k+1}$$



        and now we can differentiate $x^k$ $m$ times using the formula



        $$frac{d^m}{dx^m} x^k = (k)_m x^{k-1}$$



        where $(k)_m$ is the "falling factorial" given by $(a)_n = a(a-1)(a-2)cdots(a-n+1)$, which has $n$ factors. Note that if $m > k$, this is zero, and the derivatives thus go to zero.



        Thus



        $$frac{d^m}{dx^m} x(x - 1)^n = sum_{k=0}^{n} binom{n}{k} (-1)^{n-k} (k+1)_m x^{k+1-m}$$



        and the original is



        $$frac{d^m}{dx^m} [x log(x)] = sum_{n=1}^{infty} frac{(-1)^{n-1}}{n} left[sum_{k=0}^{n} binom{n}{k} (-1)^{n-k} (k+1)_m x^{k+1-m}right]$$



        for which a rearrangement gives



        $$frac{d^m}{dx^m} [x log(x)] = sum_{n=1}^{infty} sum_{k=0}^{n} binom{n}{k} frac{(-1)^{k+1} (k+1)_m}{n} x^{k+1-m}$$



        This series will, of course, only converge in $x in (0, 2)$, I believe, but this is the closest thing to a "closed form" you will get even though it's not technically one, but moreover you can even vary $m$ smoothly, if you use the extension $(k+1)_m = frac{Gamma(k+2)}{Gamma(k-m-2)}$ and this will vary smoothly as well, so that you will have a continuous transition from $ln x$ to $frac{1}{x}$, and thus describes, or better, interpolates, the "transition" you ask about, and uses no ad-hoc step functions.



        There is probably also some way to express the series using some kind of hypergeometric functions, but I don't feel like trying to rearrange it right now.






        share|cite|improve this answer












        So basically, you are asking for a formula that is not piecewise-defined, right?



        If so, one good way to do this is to use power series, since they are very easy to differentiate to arbitrary orders, even if the original function has a complicated mix of transcendental functions, as the terms are usually simply polynomial. To maximize the easiness, however, we might first want to make a substitution to "conjugate" the differentiation problem to a simpler one. Namely, we can use the Mercator series for $log(1 + x)$, which gives



        $$log(1 + x) = sum_{n=1}^{infty} frac{(-1)^{n-1}}{n} x^n$$



        We first rearrange this a bit, setting $x mapsto x - 1$, so that



        $$log(x) = sum_{n=1}^{infty} frac{(-1)^{n-1}}{n} (x - 1)^n$$



        thus



        $$x log(x) = sum_{n=1}^{infty} frac{(-1)^{n-1}}{n} x(x - 1)^n$$



        and so the problem now reduces to differentiating the polynomial expression $x(x - 1)^n$, i.e. finding



        $$frac{d^m}{dx^m} [x(x - 1)^n]$$



        in terms of $m$. We can expand this using the binomial theorem to get



        $$x(x - 1)^n = sum_{k=0}^{n} binom{n}{k} (-1)^{n-k} x^{k+1}$$



        and now we can differentiate $x^k$ $m$ times using the formula



        $$frac{d^m}{dx^m} x^k = (k)_m x^{k-1}$$



        where $(k)_m$ is the "falling factorial" given by $(a)_n = a(a-1)(a-2)cdots(a-n+1)$, which has $n$ factors. Note that if $m > k$, this is zero, and the derivatives thus go to zero.



        Thus



        $$frac{d^m}{dx^m} x(x - 1)^n = sum_{k=0}^{n} binom{n}{k} (-1)^{n-k} (k+1)_m x^{k+1-m}$$



        and the original is



        $$frac{d^m}{dx^m} [x log(x)] = sum_{n=1}^{infty} frac{(-1)^{n-1}}{n} left[sum_{k=0}^{n} binom{n}{k} (-1)^{n-k} (k+1)_m x^{k+1-m}right]$$



        for which a rearrangement gives



        $$frac{d^m}{dx^m} [x log(x)] = sum_{n=1}^{infty} sum_{k=0}^{n} binom{n}{k} frac{(-1)^{k+1} (k+1)_m}{n} x^{k+1-m}$$



        This series will, of course, only converge in $x in (0, 2)$, I believe, but this is the closest thing to a "closed form" you will get even though it's not technically one, but moreover you can even vary $m$ smoothly, if you use the extension $(k+1)_m = frac{Gamma(k+2)}{Gamma(k-m-2)}$ and this will vary smoothly as well, so that you will have a continuous transition from $ln x$ to $frac{1}{x}$, and thus describes, or better, interpolates, the "transition" you ask about, and uses no ad-hoc step functions.



        There is probably also some way to express the series using some kind of hypergeometric functions, but I don't feel like trying to rearrange it right now.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 19 mins ago









        The_Sympathizer

        7,1372243




        7,1372243






















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