Finding $frac{d^{n-1}}{dx^{n-1}}(1+sqrt x)^{2n}|_{x=1}$












1












$begingroup$



How can we find $$frac{d^{n-1}}{dx^{n-1}}(1+sqrt x)^{2n}Big|_{x=1},$$where $ninmathbb{N}$?




Attempt of complex integration
$$begin{aligned}&frac{d^{n-1}}{dx^{n-1}}(1+sqrt x)^{2n}Big|_{x=1}\
&=frac{(n-1)!}{2pi i}int_Cfrac{(1+sqrt z)^{2n}}{(z-1)^n}dztext{ (Cauchy's formula)}\
&=frac{(n-1)!}{pi i}int_{C'}frac{z(1+z)^{2n}}{(z^2-1)^n}dztext{ (Substitute $zto z^2$)}\
&=frac{(n-1)!}{pi i}int_{C'}frac{z(1+z)^{n}}{(z-1)^n}dz\
&=frac{(n-1)!}{pi i}int_{C''}-frac{2 (w+1) w^n}{(w-1)^3}dwtext{ (Substitute $w=frac{1+z}{z-1}$)}\
&=frac{(n-1)!}{pi i}2pi i(-2n^2)=-4ncdot n!end{aligned}$$
But the correct answer is $4ncdot n!$, I don't know where the mistake is.

Also, I wonder if there is a solution without using complex analysis.






calculus complex-analysis alternative-proof







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$endgroup$








  • 2




    $begingroup$
    The substitution $zto z^2$ could be dangerous...
    $endgroup$
    – Szeto
    Dec 10 '18 at 9:07
















1












$begingroup$



How can we find $$frac{d^{n-1}}{dx^{n-1}}(1+sqrt x)^{2n}Big|_{x=1},$$where $ninmathbb{N}$?




Attempt of complex integration
$$begin{aligned}&frac{d^{n-1}}{dx^{n-1}}(1+sqrt x)^{2n}Big|_{x=1}\
&=frac{(n-1)!}{2pi i}int_Cfrac{(1+sqrt z)^{2n}}{(z-1)^n}dztext{ (Cauchy's formula)}\
&=frac{(n-1)!}{pi i}int_{C'}frac{z(1+z)^{2n}}{(z^2-1)^n}dztext{ (Substitute $zto z^2$)}\
&=frac{(n-1)!}{pi i}int_{C'}frac{z(1+z)^{n}}{(z-1)^n}dz\
&=frac{(n-1)!}{pi i}int_{C''}-frac{2 (w+1) w^n}{(w-1)^3}dwtext{ (Substitute $w=frac{1+z}{z-1}$)}\
&=frac{(n-1)!}{pi i}2pi i(-2n^2)=-4ncdot n!end{aligned}$$
But the correct answer is $4ncdot n!$, I don't know where the mistake is.

Also, I wonder if there is a solution without using complex analysis.






calculus complex-analysis alternative-proof







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    The substitution $zto z^2$ could be dangerous...
    $endgroup$
    – Szeto
    Dec 10 '18 at 9:07














1












1








1





$begingroup$



How can we find $$frac{d^{n-1}}{dx^{n-1}}(1+sqrt x)^{2n}Big|_{x=1},$$where $ninmathbb{N}$?




Attempt of complex integration
$$begin{aligned}&frac{d^{n-1}}{dx^{n-1}}(1+sqrt x)^{2n}Big|_{x=1}\
&=frac{(n-1)!}{2pi i}int_Cfrac{(1+sqrt z)^{2n}}{(z-1)^n}dztext{ (Cauchy's formula)}\
&=frac{(n-1)!}{pi i}int_{C'}frac{z(1+z)^{2n}}{(z^2-1)^n}dztext{ (Substitute $zto z^2$)}\
&=frac{(n-1)!}{pi i}int_{C'}frac{z(1+z)^{n}}{(z-1)^n}dz\
&=frac{(n-1)!}{pi i}int_{C''}-frac{2 (w+1) w^n}{(w-1)^3}dwtext{ (Substitute $w=frac{1+z}{z-1}$)}\
&=frac{(n-1)!}{pi i}2pi i(-2n^2)=-4ncdot n!end{aligned}$$
But the correct answer is $4ncdot n!$, I don't know where the mistake is.

Also, I wonder if there is a solution without using complex analysis.






calculus complex-analysis alternative-proof







share|cite|improve this question









$endgroup$





How can we find $$frac{d^{n-1}}{dx^{n-1}}(1+sqrt x)^{2n}Big|_{x=1},$$where $ninmathbb{N}$?




Attempt of complex integration
$$begin{aligned}&frac{d^{n-1}}{dx^{n-1}}(1+sqrt x)^{2n}Big|_{x=1}\
&=frac{(n-1)!}{2pi i}int_Cfrac{(1+sqrt z)^{2n}}{(z-1)^n}dztext{ (Cauchy's formula)}\
&=frac{(n-1)!}{pi i}int_{C'}frac{z(1+z)^{2n}}{(z^2-1)^n}dztext{ (Substitute $zto z^2$)}\
&=frac{(n-1)!}{pi i}int_{C'}frac{z(1+z)^{n}}{(z-1)^n}dz\
&=frac{(n-1)!}{pi i}int_{C''}-frac{2 (w+1) w^n}{(w-1)^3}dwtext{ (Substitute $w=frac{1+z}{z-1}$)}\
&=frac{(n-1)!}{pi i}2pi i(-2n^2)=-4ncdot n!end{aligned}$$
But the correct answer is $4ncdot n!$, I don't know where the mistake is.

Also, I wonder if there is a solution without using complex analysis.






calculus complex-analysis alternative-proof




calculus complex-analysis alternative-proof






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asked Dec 10 '18 at 8:31









Kemono ChenKemono Chen

3,0521743




3,0521743








  • 2




    $begingroup$
    The substitution $zto z^2$ could be dangerous...
    $endgroup$
    – Szeto
    Dec 10 '18 at 9:07














  • 2




    $begingroup$
    The substitution $zto z^2$ could be dangerous...
    $endgroup$
    – Szeto
    Dec 10 '18 at 9:07








2




2




$begingroup$
The substitution $zto z^2$ could be dangerous...
$endgroup$
– Szeto
Dec 10 '18 at 9:07




$begingroup$
The substitution $zto z^2$ could be dangerous...
$endgroup$
– Szeto
Dec 10 '18 at 9:07










1 Answer
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$begingroup$

The transformation from $z$ to $w$ reverses the orientation of the closed curve. Therefore, you need an extra minus sign, leading to the correct result.



Alternatively, let $u:=z-1$, so you will not need to worry about the orientation of the curve, and the required result is
$$frac{(n-1)!}{pi text{i}},oint_{C'''},frac{(u+1)(u+2)^n}{u^n},text{d}u$$
for some positively oriented closed curve $C'''$ about the point $u=0$. Since
$$(u+1)(u+2)^n=u^{n+1}+2n,u^{n}+2n^2,u^{n-1}+mathcal{O}(u^{n-2}),,$$
we get the required result.






share|cite|improve this answer









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    $begingroup$

    The transformation from $z$ to $w$ reverses the orientation of the closed curve. Therefore, you need an extra minus sign, leading to the correct result.



    Alternatively, let $u:=z-1$, so you will not need to worry about the orientation of the curve, and the required result is
    $$frac{(n-1)!}{pi text{i}},oint_{C'''},frac{(u+1)(u+2)^n}{u^n},text{d}u$$
    for some positively oriented closed curve $C'''$ about the point $u=0$. Since
    $$(u+1)(u+2)^n=u^{n+1}+2n,u^{n}+2n^2,u^{n-1}+mathcal{O}(u^{n-2}),,$$
    we get the required result.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      The transformation from $z$ to $w$ reverses the orientation of the closed curve. Therefore, you need an extra minus sign, leading to the correct result.



      Alternatively, let $u:=z-1$, so you will not need to worry about the orientation of the curve, and the required result is
      $$frac{(n-1)!}{pi text{i}},oint_{C'''},frac{(u+1)(u+2)^n}{u^n},text{d}u$$
      for some positively oriented closed curve $C'''$ about the point $u=0$. Since
      $$(u+1)(u+2)^n=u^{n+1}+2n,u^{n}+2n^2,u^{n-1}+mathcal{O}(u^{n-2}),,$$
      we get the required result.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        The transformation from $z$ to $w$ reverses the orientation of the closed curve. Therefore, you need an extra minus sign, leading to the correct result.



        Alternatively, let $u:=z-1$, so you will not need to worry about the orientation of the curve, and the required result is
        $$frac{(n-1)!}{pi text{i}},oint_{C'''},frac{(u+1)(u+2)^n}{u^n},text{d}u$$
        for some positively oriented closed curve $C'''$ about the point $u=0$. Since
        $$(u+1)(u+2)^n=u^{n+1}+2n,u^{n}+2n^2,u^{n-1}+mathcal{O}(u^{n-2}),,$$
        we get the required result.






        share|cite|improve this answer









        $endgroup$



        The transformation from $z$ to $w$ reverses the orientation of the closed curve. Therefore, you need an extra minus sign, leading to the correct result.



        Alternatively, let $u:=z-1$, so you will not need to worry about the orientation of the curve, and the required result is
        $$frac{(n-1)!}{pi text{i}},oint_{C'''},frac{(u+1)(u+2)^n}{u^n},text{d}u$$
        for some positively oriented closed curve $C'''$ about the point $u=0$. Since
        $$(u+1)(u+2)^n=u^{n+1}+2n,u^{n}+2n^2,u^{n-1}+mathcal{O}(u^{n-2}),,$$
        we get the required result.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 10 '18 at 11:26









        BatominovskiBatominovski

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