Maclaurin expansion of $arctan(x)/(1 − x).$












10












$begingroup$


How was this Maclaurin expansion derived? For each $|x|<1,$
begin{align} left( frac{arctan(x)}{1-x} right)&=left( sum^{infty}_{k=0}x^kright)left(sum^{infty}_{j=0}dfrac{(-1)^j x^{2j+1}}{2j+1}right)\&= sum^{infty}_{k=0}left(sum_{jin D_k} dfrac{ (-1)^{j} }{2j+1}right)x^k,;text{where}; D_k={jin Bbb{N}:0leq jleq (k-1)/2}.end{align}



HERE'S MY TRIAL



begin{align} left( frac{arctan(x)}{1-x} right)&=left( sum^{infty}_{k=0}x^kright)left(sum^{infty}_{j=0}dfrac{(-1)^j x^{2j+1}}{2j+1}right)\&= sum^{infty}_{k=0}left(sum^{k}_{j=0} x^jdfrac{ (-1)^{(k-j)} x^{2(k-j)+1}}{2(k-j)+1}right),;text{for}; kin Bbb{N}\&stackrel{text{how?}}{=} sum^{infty}_{k=0}left(sum_{jin D_k} dfrac{ (-1)^{j} }{2j+1}right)x^k.end{align}






real-analysis calculus sequences-and-series analysis taylor-expansion







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$endgroup$








  • 1




    $begingroup$
    I'm not sure what you're asking here. Do you want a different expansion? This is a fairly straightforward approach.
    $endgroup$
    – robjohn♦
    Jan 7 at 15:36










  • $begingroup$
    @robjohn: I have edited it. Thanks.
    $endgroup$
    – Omojola Micheal
    Jan 7 at 16:02






  • 1




    $begingroup$
    Look up the Cauchy product.
    $endgroup$
    – Darks
    Jan 7 at 16:30
















10












$begingroup$


How was this Maclaurin expansion derived? For each $|x|<1,$
begin{align} left( frac{arctan(x)}{1-x} right)&=left( sum^{infty}_{k=0}x^kright)left(sum^{infty}_{j=0}dfrac{(-1)^j x^{2j+1}}{2j+1}right)\&= sum^{infty}_{k=0}left(sum_{jin D_k} dfrac{ (-1)^{j} }{2j+1}right)x^k,;text{where}; D_k={jin Bbb{N}:0leq jleq (k-1)/2}.end{align}



HERE'S MY TRIAL



begin{align} left( frac{arctan(x)}{1-x} right)&=left( sum^{infty}_{k=0}x^kright)left(sum^{infty}_{j=0}dfrac{(-1)^j x^{2j+1}}{2j+1}right)\&= sum^{infty}_{k=0}left(sum^{k}_{j=0} x^jdfrac{ (-1)^{(k-j)} x^{2(k-j)+1}}{2(k-j)+1}right),;text{for}; kin Bbb{N}\&stackrel{text{how?}}{=} sum^{infty}_{k=0}left(sum_{jin D_k} dfrac{ (-1)^{j} }{2j+1}right)x^k.end{align}






real-analysis calculus sequences-and-series analysis taylor-expansion







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$endgroup$








  • 1




    $begingroup$
    I'm not sure what you're asking here. Do you want a different expansion? This is a fairly straightforward approach.
    $endgroup$
    – robjohn♦
    Jan 7 at 15:36










  • $begingroup$
    @robjohn: I have edited it. Thanks.
    $endgroup$
    – Omojola Micheal
    Jan 7 at 16:02






  • 1




    $begingroup$
    Look up the Cauchy product.
    $endgroup$
    – Darks
    Jan 7 at 16:30














10












10








10


4



$begingroup$


How was this Maclaurin expansion derived? For each $|x|<1,$
begin{align} left( frac{arctan(x)}{1-x} right)&=left( sum^{infty}_{k=0}x^kright)left(sum^{infty}_{j=0}dfrac{(-1)^j x^{2j+1}}{2j+1}right)\&= sum^{infty}_{k=0}left(sum_{jin D_k} dfrac{ (-1)^{j} }{2j+1}right)x^k,;text{where}; D_k={jin Bbb{N}:0leq jleq (k-1)/2}.end{align}



HERE'S MY TRIAL



begin{align} left( frac{arctan(x)}{1-x} right)&=left( sum^{infty}_{k=0}x^kright)left(sum^{infty}_{j=0}dfrac{(-1)^j x^{2j+1}}{2j+1}right)\&= sum^{infty}_{k=0}left(sum^{k}_{j=0} x^jdfrac{ (-1)^{(k-j)} x^{2(k-j)+1}}{2(k-j)+1}right),;text{for}; kin Bbb{N}\&stackrel{text{how?}}{=} sum^{infty}_{k=0}left(sum_{jin D_k} dfrac{ (-1)^{j} }{2j+1}right)x^k.end{align}






real-analysis calculus sequences-and-series analysis taylor-expansion







share|cite|improve this question











$endgroup$




How was this Maclaurin expansion derived? For each $|x|<1,$
begin{align} left( frac{arctan(x)}{1-x} right)&=left( sum^{infty}_{k=0}x^kright)left(sum^{infty}_{j=0}dfrac{(-1)^j x^{2j+1}}{2j+1}right)\&= sum^{infty}_{k=0}left(sum_{jin D_k} dfrac{ (-1)^{j} }{2j+1}right)x^k,;text{where}; D_k={jin Bbb{N}:0leq jleq (k-1)/2}.end{align}



HERE'S MY TRIAL



begin{align} left( frac{arctan(x)}{1-x} right)&=left( sum^{infty}_{k=0}x^kright)left(sum^{infty}_{j=0}dfrac{(-1)^j x^{2j+1}}{2j+1}right)\&= sum^{infty}_{k=0}left(sum^{k}_{j=0} x^jdfrac{ (-1)^{(k-j)} x^{2(k-j)+1}}{2(k-j)+1}right),;text{for}; kin Bbb{N}\&stackrel{text{how?}}{=} sum^{infty}_{k=0}left(sum_{jin D_k} dfrac{ (-1)^{j} }{2j+1}right)x^k.end{align}






real-analysis calculus sequences-and-series analysis taylor-expansion




real-analysis calculus sequences-and-series analysis taylor-expansion






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edited Jan 7 at 22:49









Matemáticos Chibchas

5,44322141




5,44322141










asked Jan 7 at 14:33









Omojola MichealOmojola Micheal

2,064424




2,064424








  • 1




    $begingroup$
    I'm not sure what you're asking here. Do you want a different expansion? This is a fairly straightforward approach.
    $endgroup$
    – robjohn♦
    Jan 7 at 15:36










  • $begingroup$
    @robjohn: I have edited it. Thanks.
    $endgroup$
    – Omojola Micheal
    Jan 7 at 16:02






  • 1




    $begingroup$
    Look up the Cauchy product.
    $endgroup$
    – Darks
    Jan 7 at 16:30














  • 1




    $begingroup$
    I'm not sure what you're asking here. Do you want a different expansion? This is a fairly straightforward approach.
    $endgroup$
    – robjohn♦
    Jan 7 at 15:36










  • $begingroup$
    @robjohn: I have edited it. Thanks.
    $endgroup$
    – Omojola Micheal
    Jan 7 at 16:02






  • 1




    $begingroup$
    Look up the Cauchy product.
    $endgroup$
    – Darks
    Jan 7 at 16:30








1




1




$begingroup$
I'm not sure what you're asking here. Do you want a different expansion? This is a fairly straightforward approach.
$endgroup$
– robjohn♦
Jan 7 at 15:36




$begingroup$
I'm not sure what you're asking here. Do you want a different expansion? This is a fairly straightforward approach.
$endgroup$
– robjohn♦
Jan 7 at 15:36












$begingroup$
@robjohn: I have edited it. Thanks.
$endgroup$
– Omojola Micheal
Jan 7 at 16:02




$begingroup$
@robjohn: I have edited it. Thanks.
$endgroup$
– Omojola Micheal
Jan 7 at 16:02




1




1




$begingroup$
Look up the Cauchy product.
$endgroup$
– Darks
Jan 7 at 16:30




$begingroup$
Look up the Cauchy product.
$endgroup$
– Darks
Jan 7 at 16:30










2 Answers
2






active

oldest

votes


















3












$begingroup$

The idea here is to use the Cauchy Product. However, since one series has exponents of $k$ and the other series has exponents of $2j+1$, we need to extract the essence of the product formula: that is, for a given $m$, the products of which terms give $x^m$? That would be when $k+2j+1=m$. Thus, the coefficient of $x^m$ in the final product is
$$
sum_{j=0}^{leftlfloorfrac{m-1}2rightrfloor}overbrace{vphantom{b} a_{m-2j-1} }^{substack{text{coefficient}\text{of $x^{m-2j-1}$}}}overbrace{ b_j }^{substack{text{coefficient}\text{of $x^{2j+1}$}}}
$$
That is,
$$
sum_{k=0}^infty a_kx^ksum_{j=0}^infty b_jx^{2j+1}
=sum_{m=0}^inftysum_{j=0}^{leftlfloorfrac{m-1}2rightrfloor}a_{m-2j-1}b_jx^m
$$
Since $a_k=1$ for all $kge0$, we have
$$
sum_{m=0}^inftysum_{j=0}^{leftlfloorfrac{m-1}2rightrfloor}overbrace{ frac{(-1)^j}{2j+1} }^{b_j}x^m
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    (+1) Thanks for the correction.
    $endgroup$
    – Omojola Micheal
    Jan 8 at 5:35



















1












$begingroup$

Corrected: Thanks to robjohn for the mentorship. Since
begin{align} left( sum^{infty}_{k=0}a_k x^kright)left( sum^{infty}_{j=0}b_j x^jright)&=sum^{infty}_{r=0}left( sum^{r}_{j=0}b_j a_{r-j} right)x^{r},;text{where}; r=k+j;text{and};rinBbb{N},.end{align}
we have that,



begin{align} left( frac{arctan x }{1-x} right)&=left( sum^{infty}_{k=0}a_k x^kright)left( sum^{infty}_{j=0}b_j x^{2j+1}right)\&=sum^{infty}_{gamma=0}left( sum^{frac{gamma-1}{2}}_{j=0}b_j a_{gamma-2j-1} right)x^{gamma},;text{where}; gamma=k+2j+1;text{and};gammainBbb{N},\&=sum^{infty}_{gamma=0}left( sum^{frac{gamma-1}{2}}_{j=0}dfrac{ (-1)^{j} }{2j+1} right)x^{gamma},\&= sum^{infty}_{gamma=0}left(sum_{jin D_gamma} dfrac{ (-1)^{j} }{2j+1}right)x^gamma.end{align}
where $D_gamma={jin Bbb{N}:0leq jleq (gamma-1)/2},;a_{gamma-2j-1} =1,,jin D_gamma ,;b_j= (-1)^{j} /(2j+1).$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    In the last set of equations, the second line is wrong for exponents $k$ and $2j+1$, and the upper limit in the inner sum on the third line does not make sense because $j$ is the variable of iteration. It is not obvious how $frac{gamma-1}2$ appears in the fourth line (especially since that line says $gamma=k+2j+1$ which is the upper limit in the prevous line).
    $endgroup$
    – robjohn♦
    Jan 7 at 23:29










  • $begingroup$
    @robjohn: Will fix it.
    $endgroup$
    – Omojola Micheal
    Jan 7 at 23:30










  • $begingroup$
    @robjohn: I have fixed it. Thank you!
    $endgroup$
    – Omojola Micheal
    Jan 8 at 5:20












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

The idea here is to use the Cauchy Product. However, since one series has exponents of $k$ and the other series has exponents of $2j+1$, we need to extract the essence of the product formula: that is, for a given $m$, the products of which terms give $x^m$? That would be when $k+2j+1=m$. Thus, the coefficient of $x^m$ in the final product is
$$
sum_{j=0}^{leftlfloorfrac{m-1}2rightrfloor}overbrace{vphantom{b} a_{m-2j-1} }^{substack{text{coefficient}\text{of $x^{m-2j-1}$}}}overbrace{ b_j }^{substack{text{coefficient}\text{of $x^{2j+1}$}}}
$$
That is,
$$
sum_{k=0}^infty a_kx^ksum_{j=0}^infty b_jx^{2j+1}
=sum_{m=0}^inftysum_{j=0}^{leftlfloorfrac{m-1}2rightrfloor}a_{m-2j-1}b_jx^m
$$
Since $a_k=1$ for all $kge0$, we have
$$
sum_{m=0}^inftysum_{j=0}^{leftlfloorfrac{m-1}2rightrfloor}overbrace{ frac{(-1)^j}{2j+1} }^{b_j}x^m
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    (+1) Thanks for the correction.
    $endgroup$
    – Omojola Micheal
    Jan 8 at 5:35
















3












$begingroup$

The idea here is to use the Cauchy Product. However, since one series has exponents of $k$ and the other series has exponents of $2j+1$, we need to extract the essence of the product formula: that is, for a given $m$, the products of which terms give $x^m$? That would be when $k+2j+1=m$. Thus, the coefficient of $x^m$ in the final product is
$$
sum_{j=0}^{leftlfloorfrac{m-1}2rightrfloor}overbrace{vphantom{b} a_{m-2j-1} }^{substack{text{coefficient}\text{of $x^{m-2j-1}$}}}overbrace{ b_j }^{substack{text{coefficient}\text{of $x^{2j+1}$}}}
$$
That is,
$$
sum_{k=0}^infty a_kx^ksum_{j=0}^infty b_jx^{2j+1}
=sum_{m=0}^inftysum_{j=0}^{leftlfloorfrac{m-1}2rightrfloor}a_{m-2j-1}b_jx^m
$$
Since $a_k=1$ for all $kge0$, we have
$$
sum_{m=0}^inftysum_{j=0}^{leftlfloorfrac{m-1}2rightrfloor}overbrace{ frac{(-1)^j}{2j+1} }^{b_j}x^m
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    (+1) Thanks for the correction.
    $endgroup$
    – Omojola Micheal
    Jan 8 at 5:35














3












3








3





$begingroup$

The idea here is to use the Cauchy Product. However, since one series has exponents of $k$ and the other series has exponents of $2j+1$, we need to extract the essence of the product formula: that is, for a given $m$, the products of which terms give $x^m$? That would be when $k+2j+1=m$. Thus, the coefficient of $x^m$ in the final product is
$$
sum_{j=0}^{leftlfloorfrac{m-1}2rightrfloor}overbrace{vphantom{b} a_{m-2j-1} }^{substack{text{coefficient}\text{of $x^{m-2j-1}$}}}overbrace{ b_j }^{substack{text{coefficient}\text{of $x^{2j+1}$}}}
$$
That is,
$$
sum_{k=0}^infty a_kx^ksum_{j=0}^infty b_jx^{2j+1}
=sum_{m=0}^inftysum_{j=0}^{leftlfloorfrac{m-1}2rightrfloor}a_{m-2j-1}b_jx^m
$$
Since $a_k=1$ for all $kge0$, we have
$$
sum_{m=0}^inftysum_{j=0}^{leftlfloorfrac{m-1}2rightrfloor}overbrace{ frac{(-1)^j}{2j+1} }^{b_j}x^m
$$






share|cite|improve this answer









$endgroup$



The idea here is to use the Cauchy Product. However, since one series has exponents of $k$ and the other series has exponents of $2j+1$, we need to extract the essence of the product formula: that is, for a given $m$, the products of which terms give $x^m$? That would be when $k+2j+1=m$. Thus, the coefficient of $x^m$ in the final product is
$$
sum_{j=0}^{leftlfloorfrac{m-1}2rightrfloor}overbrace{vphantom{b} a_{m-2j-1} }^{substack{text{coefficient}\text{of $x^{m-2j-1}$}}}overbrace{ b_j }^{substack{text{coefficient}\text{of $x^{2j+1}$}}}
$$
That is,
$$
sum_{k=0}^infty a_kx^ksum_{j=0}^infty b_jx^{2j+1}
=sum_{m=0}^inftysum_{j=0}^{leftlfloorfrac{m-1}2rightrfloor}a_{m-2j-1}b_jx^m
$$
Since $a_k=1$ for all $kge0$, we have
$$
sum_{m=0}^inftysum_{j=0}^{leftlfloorfrac{m-1}2rightrfloor}overbrace{ frac{(-1)^j}{2j+1} }^{b_j}x^m
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 7 at 17:36









robjohn♦robjohn

271k27313642




271k27313642












  • $begingroup$
    (+1) Thanks for the correction.
    $endgroup$
    – Omojola Micheal
    Jan 8 at 5:35


















  • $begingroup$
    (+1) Thanks for the correction.
    $endgroup$
    – Omojola Micheal
    Jan 8 at 5:35
















$begingroup$
(+1) Thanks for the correction.
$endgroup$
– Omojola Micheal
Jan 8 at 5:35




$begingroup$
(+1) Thanks for the correction.
$endgroup$
– Omojola Micheal
Jan 8 at 5:35











1












$begingroup$

Corrected: Thanks to robjohn for the mentorship. Since
begin{align} left( sum^{infty}_{k=0}a_k x^kright)left( sum^{infty}_{j=0}b_j x^jright)&=sum^{infty}_{r=0}left( sum^{r}_{j=0}b_j a_{r-j} right)x^{r},;text{where}; r=k+j;text{and};rinBbb{N},.end{align}
we have that,



begin{align} left( frac{arctan x }{1-x} right)&=left( sum^{infty}_{k=0}a_k x^kright)left( sum^{infty}_{j=0}b_j x^{2j+1}right)\&=sum^{infty}_{gamma=0}left( sum^{frac{gamma-1}{2}}_{j=0}b_j a_{gamma-2j-1} right)x^{gamma},;text{where}; gamma=k+2j+1;text{and};gammainBbb{N},\&=sum^{infty}_{gamma=0}left( sum^{frac{gamma-1}{2}}_{j=0}dfrac{ (-1)^{j} }{2j+1} right)x^{gamma},\&= sum^{infty}_{gamma=0}left(sum_{jin D_gamma} dfrac{ (-1)^{j} }{2j+1}right)x^gamma.end{align}
where $D_gamma={jin Bbb{N}:0leq jleq (gamma-1)/2},;a_{gamma-2j-1} =1,,jin D_gamma ,;b_j= (-1)^{j} /(2j+1).$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    In the last set of equations, the second line is wrong for exponents $k$ and $2j+1$, and the upper limit in the inner sum on the third line does not make sense because $j$ is the variable of iteration. It is not obvious how $frac{gamma-1}2$ appears in the fourth line (especially since that line says $gamma=k+2j+1$ which is the upper limit in the prevous line).
    $endgroup$
    – robjohn♦
    Jan 7 at 23:29










  • $begingroup$
    @robjohn: Will fix it.
    $endgroup$
    – Omojola Micheal
    Jan 7 at 23:30










  • $begingroup$
    @robjohn: I have fixed it. Thank you!
    $endgroup$
    – Omojola Micheal
    Jan 8 at 5:20
















1












$begingroup$

Corrected: Thanks to robjohn for the mentorship. Since
begin{align} left( sum^{infty}_{k=0}a_k x^kright)left( sum^{infty}_{j=0}b_j x^jright)&=sum^{infty}_{r=0}left( sum^{r}_{j=0}b_j a_{r-j} right)x^{r},;text{where}; r=k+j;text{and};rinBbb{N},.end{align}
we have that,



begin{align} left( frac{arctan x }{1-x} right)&=left( sum^{infty}_{k=0}a_k x^kright)left( sum^{infty}_{j=0}b_j x^{2j+1}right)\&=sum^{infty}_{gamma=0}left( sum^{frac{gamma-1}{2}}_{j=0}b_j a_{gamma-2j-1} right)x^{gamma},;text{where}; gamma=k+2j+1;text{and};gammainBbb{N},\&=sum^{infty}_{gamma=0}left( sum^{frac{gamma-1}{2}}_{j=0}dfrac{ (-1)^{j} }{2j+1} right)x^{gamma},\&= sum^{infty}_{gamma=0}left(sum_{jin D_gamma} dfrac{ (-1)^{j} }{2j+1}right)x^gamma.end{align}
where $D_gamma={jin Bbb{N}:0leq jleq (gamma-1)/2},;a_{gamma-2j-1} =1,,jin D_gamma ,;b_j= (-1)^{j} /(2j+1).$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    In the last set of equations, the second line is wrong for exponents $k$ and $2j+1$, and the upper limit in the inner sum on the third line does not make sense because $j$ is the variable of iteration. It is not obvious how $frac{gamma-1}2$ appears in the fourth line (especially since that line says $gamma=k+2j+1$ which is the upper limit in the prevous line).
    $endgroup$
    – robjohn♦
    Jan 7 at 23:29










  • $begingroup$
    @robjohn: Will fix it.
    $endgroup$
    – Omojola Micheal
    Jan 7 at 23:30










  • $begingroup$
    @robjohn: I have fixed it. Thank you!
    $endgroup$
    – Omojola Micheal
    Jan 8 at 5:20














1












1








1





$begingroup$

Corrected: Thanks to robjohn for the mentorship. Since
begin{align} left( sum^{infty}_{k=0}a_k x^kright)left( sum^{infty}_{j=0}b_j x^jright)&=sum^{infty}_{r=0}left( sum^{r}_{j=0}b_j a_{r-j} right)x^{r},;text{where}; r=k+j;text{and};rinBbb{N},.end{align}
we have that,



begin{align} left( frac{arctan x }{1-x} right)&=left( sum^{infty}_{k=0}a_k x^kright)left( sum^{infty}_{j=0}b_j x^{2j+1}right)\&=sum^{infty}_{gamma=0}left( sum^{frac{gamma-1}{2}}_{j=0}b_j a_{gamma-2j-1} right)x^{gamma},;text{where}; gamma=k+2j+1;text{and};gammainBbb{N},\&=sum^{infty}_{gamma=0}left( sum^{frac{gamma-1}{2}}_{j=0}dfrac{ (-1)^{j} }{2j+1} right)x^{gamma},\&= sum^{infty}_{gamma=0}left(sum_{jin D_gamma} dfrac{ (-1)^{j} }{2j+1}right)x^gamma.end{align}
where $D_gamma={jin Bbb{N}:0leq jleq (gamma-1)/2},;a_{gamma-2j-1} =1,,jin D_gamma ,;b_j= (-1)^{j} /(2j+1).$






share|cite|improve this answer











$endgroup$



Corrected: Thanks to robjohn for the mentorship. Since
begin{align} left( sum^{infty}_{k=0}a_k x^kright)left( sum^{infty}_{j=0}b_j x^jright)&=sum^{infty}_{r=0}left( sum^{r}_{j=0}b_j a_{r-j} right)x^{r},;text{where}; r=k+j;text{and};rinBbb{N},.end{align}
we have that,



begin{align} left( frac{arctan x }{1-x} right)&=left( sum^{infty}_{k=0}a_k x^kright)left( sum^{infty}_{j=0}b_j x^{2j+1}right)\&=sum^{infty}_{gamma=0}left( sum^{frac{gamma-1}{2}}_{j=0}b_j a_{gamma-2j-1} right)x^{gamma},;text{where}; gamma=k+2j+1;text{and};gammainBbb{N},\&=sum^{infty}_{gamma=0}left( sum^{frac{gamma-1}{2}}_{j=0}dfrac{ (-1)^{j} }{2j+1} right)x^{gamma},\&= sum^{infty}_{gamma=0}left(sum_{jin D_gamma} dfrac{ (-1)^{j} }{2j+1}right)x^gamma.end{align}
where $D_gamma={jin Bbb{N}:0leq jleq (gamma-1)/2},;a_{gamma-2j-1} =1,,jin D_gamma ,;b_j= (-1)^{j} /(2j+1).$







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edited Jan 8 at 11:30

























answered Jan 7 at 21:57









Omojola MichealOmojola Micheal

2,064424




2,064424












  • $begingroup$
    In the last set of equations, the second line is wrong for exponents $k$ and $2j+1$, and the upper limit in the inner sum on the third line does not make sense because $j$ is the variable of iteration. It is not obvious how $frac{gamma-1}2$ appears in the fourth line (especially since that line says $gamma=k+2j+1$ which is the upper limit in the prevous line).
    $endgroup$
    – robjohn♦
    Jan 7 at 23:29










  • $begingroup$
    @robjohn: Will fix it.
    $endgroup$
    – Omojola Micheal
    Jan 7 at 23:30










  • $begingroup$
    @robjohn: I have fixed it. Thank you!
    $endgroup$
    – Omojola Micheal
    Jan 8 at 5:20


















  • $begingroup$
    In the last set of equations, the second line is wrong for exponents $k$ and $2j+1$, and the upper limit in the inner sum on the third line does not make sense because $j$ is the variable of iteration. It is not obvious how $frac{gamma-1}2$ appears in the fourth line (especially since that line says $gamma=k+2j+1$ which is the upper limit in the prevous line).
    $endgroup$
    – robjohn♦
    Jan 7 at 23:29










  • $begingroup$
    @robjohn: Will fix it.
    $endgroup$
    – Omojola Micheal
    Jan 7 at 23:30










  • $begingroup$
    @robjohn: I have fixed it. Thank you!
    $endgroup$
    – Omojola Micheal
    Jan 8 at 5:20
















$begingroup$
In the last set of equations, the second line is wrong for exponents $k$ and $2j+1$, and the upper limit in the inner sum on the third line does not make sense because $j$ is the variable of iteration. It is not obvious how $frac{gamma-1}2$ appears in the fourth line (especially since that line says $gamma=k+2j+1$ which is the upper limit in the prevous line).
$endgroup$
– robjohn♦
Jan 7 at 23:29




$begingroup$
In the last set of equations, the second line is wrong for exponents $k$ and $2j+1$, and the upper limit in the inner sum on the third line does not make sense because $j$ is the variable of iteration. It is not obvious how $frac{gamma-1}2$ appears in the fourth line (especially since that line says $gamma=k+2j+1$ which is the upper limit in the prevous line).
$endgroup$
– robjohn♦
Jan 7 at 23:29












$begingroup$
@robjohn: Will fix it.
$endgroup$
– Omojola Micheal
Jan 7 at 23:30




$begingroup$
@robjohn: Will fix it.
$endgroup$
– Omojola Micheal
Jan 7 at 23:30












$begingroup$
@robjohn: I have fixed it. Thank you!
$endgroup$
– Omojola Micheal
Jan 8 at 5:20




$begingroup$
@robjohn: I have fixed it. Thank you!
$endgroup$
– Omojola Micheal
Jan 8 at 5:20


















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