Writing a complex function as a power series?












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$begingroup$


I have been asked to write the following summation as a power series:



$$sum_{n geq 500} i^n frac {z^{5n-2}}{n!}. $$



I know that by comparison to the power series $$sum_{n geq 0} a_n (z-a)^n, $$ we can let $$ a_n = frac {i^n}{n!},$$ and we can let $$a=0.$$ I am unsure how to represent the $$ n geq 500 text { and the power } 5n-2. $$



Thanks






complex-analysis power-series







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  • $begingroup$
    $i^n z^{5n-2}=frac{1}{z^2}(iz^5)^n$
    $endgroup$
    – Tito Eliatron
    Dec 2 '18 at 17:23


















0












$begingroup$


I have been asked to write the following summation as a power series:



$$sum_{n geq 500} i^n frac {z^{5n-2}}{n!}. $$



I know that by comparison to the power series $$sum_{n geq 0} a_n (z-a)^n, $$ we can let $$ a_n = frac {i^n}{n!},$$ and we can let $$a=0.$$ I am unsure how to represent the $$ n geq 500 text { and the power } 5n-2. $$



Thanks






complex-analysis power-series







share|cite|improve this question









$endgroup$












  • $begingroup$
    $i^n z^{5n-2}=frac{1}{z^2}(iz^5)^n$
    $endgroup$
    – Tito Eliatron
    Dec 2 '18 at 17:23
















0












0








0


0



$begingroup$


I have been asked to write the following summation as a power series:



$$sum_{n geq 500} i^n frac {z^{5n-2}}{n!}. $$



I know that by comparison to the power series $$sum_{n geq 0} a_n (z-a)^n, $$ we can let $$ a_n = frac {i^n}{n!},$$ and we can let $$a=0.$$ I am unsure how to represent the $$ n geq 500 text { and the power } 5n-2. $$



Thanks






complex-analysis power-series







share|cite|improve this question









$endgroup$




I have been asked to write the following summation as a power series:



$$sum_{n geq 500} i^n frac {z^{5n-2}}{n!}. $$



I know that by comparison to the power series $$sum_{n geq 0} a_n (z-a)^n, $$ we can let $$ a_n = frac {i^n}{n!},$$ and we can let $$a=0.$$ I am unsure how to represent the $$ n geq 500 text { and the power } 5n-2. $$



Thanks






complex-analysis power-series




complex-analysis power-series






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share|cite|improve this question











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asked Dec 2 '18 at 17:21









YalovetoseeitYalovetoseeit

92




92












  • $begingroup$
    $i^n z^{5n-2}=frac{1}{z^2}(iz^5)^n$
    $endgroup$
    – Tito Eliatron
    Dec 2 '18 at 17:23




















  • $begingroup$
    $i^n z^{5n-2}=frac{1}{z^2}(iz^5)^n$
    $endgroup$
    – Tito Eliatron
    Dec 2 '18 at 17:23


















$begingroup$
$i^n z^{5n-2}=frac{1}{z^2}(iz^5)^n$
$endgroup$
– Tito Eliatron
Dec 2 '18 at 17:23






$begingroup$
$i^n z^{5n-2}=frac{1}{z^2}(iz^5)^n$
$endgroup$
– Tito Eliatron
Dec 2 '18 at 17:23












1 Answer
1






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oldest

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0












$begingroup$

Define$$a_n=begin{cases}frac{i^{frac{n+2}5}}{left(frac{n+2}5right)!}&text{ if }5mid n+2text{ and }ngeqslant2,498\0&text{ otherwise.}end{cases}$$






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$endgroup$













  • $begingroup$
    What exactly do you mean by if 5 | n+2 ?
    $endgroup$
    – Yalovetoseeit
    Dec 2 '18 at 20:16










  • $begingroup$
    It means that $n+2$ is a multiple of $5$.
    $endgroup$
    – José Carlos Santos
    Dec 2 '18 at 20:19










  • $begingroup$
    Oh okay I wasn't even thinking about that function I thought it was a typo! Thanks
    $endgroup$
    – Yalovetoseeit
    Dec 2 '18 at 20:21











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1 Answer
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1 Answer
1






active

oldest

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active

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0












$begingroup$

Define$$a_n=begin{cases}frac{i^{frac{n+2}5}}{left(frac{n+2}5right)!}&text{ if }5mid n+2text{ and }ngeqslant2,498\0&text{ otherwise.}end{cases}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What exactly do you mean by if 5 | n+2 ?
    $endgroup$
    – Yalovetoseeit
    Dec 2 '18 at 20:16










  • $begingroup$
    It means that $n+2$ is a multiple of $5$.
    $endgroup$
    – José Carlos Santos
    Dec 2 '18 at 20:19










  • $begingroup$
    Oh okay I wasn't even thinking about that function I thought it was a typo! Thanks
    $endgroup$
    – Yalovetoseeit
    Dec 2 '18 at 20:21
















0












$begingroup$

Define$$a_n=begin{cases}frac{i^{frac{n+2}5}}{left(frac{n+2}5right)!}&text{ if }5mid n+2text{ and }ngeqslant2,498\0&text{ otherwise.}end{cases}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What exactly do you mean by if 5 | n+2 ?
    $endgroup$
    – Yalovetoseeit
    Dec 2 '18 at 20:16










  • $begingroup$
    It means that $n+2$ is a multiple of $5$.
    $endgroup$
    – José Carlos Santos
    Dec 2 '18 at 20:19










  • $begingroup$
    Oh okay I wasn't even thinking about that function I thought it was a typo! Thanks
    $endgroup$
    – Yalovetoseeit
    Dec 2 '18 at 20:21














0












0








0





$begingroup$

Define$$a_n=begin{cases}frac{i^{frac{n+2}5}}{left(frac{n+2}5right)!}&text{ if }5mid n+2text{ and }ngeqslant2,498\0&text{ otherwise.}end{cases}$$






share|cite|improve this answer









$endgroup$



Define$$a_n=begin{cases}frac{i^{frac{n+2}5}}{left(frac{n+2}5right)!}&text{ if }5mid n+2text{ and }ngeqslant2,498\0&text{ otherwise.}end{cases}$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 2 '18 at 17:26









José Carlos SantosJosé Carlos Santos

154k22124227




154k22124227












  • $begingroup$
    What exactly do you mean by if 5 | n+2 ?
    $endgroup$
    – Yalovetoseeit
    Dec 2 '18 at 20:16










  • $begingroup$
    It means that $n+2$ is a multiple of $5$.
    $endgroup$
    – José Carlos Santos
    Dec 2 '18 at 20:19










  • $begingroup$
    Oh okay I wasn't even thinking about that function I thought it was a typo! Thanks
    $endgroup$
    – Yalovetoseeit
    Dec 2 '18 at 20:21


















  • $begingroup$
    What exactly do you mean by if 5 | n+2 ?
    $endgroup$
    – Yalovetoseeit
    Dec 2 '18 at 20:16










  • $begingroup$
    It means that $n+2$ is a multiple of $5$.
    $endgroup$
    – José Carlos Santos
    Dec 2 '18 at 20:19










  • $begingroup$
    Oh okay I wasn't even thinking about that function I thought it was a typo! Thanks
    $endgroup$
    – Yalovetoseeit
    Dec 2 '18 at 20:21
















$begingroup$
What exactly do you mean by if 5 | n+2 ?
$endgroup$
– Yalovetoseeit
Dec 2 '18 at 20:16




$begingroup$
What exactly do you mean by if 5 | n+2 ?
$endgroup$
– Yalovetoseeit
Dec 2 '18 at 20:16












$begingroup$
It means that $n+2$ is a multiple of $5$.
$endgroup$
– José Carlos Santos
Dec 2 '18 at 20:19




$begingroup$
It means that $n+2$ is a multiple of $5$.
$endgroup$
– José Carlos Santos
Dec 2 '18 at 20:19












$begingroup$
Oh okay I wasn't even thinking about that function I thought it was a typo! Thanks
$endgroup$
– Yalovetoseeit
Dec 2 '18 at 20:21




$begingroup$
Oh okay I wasn't even thinking about that function I thought it was a typo! Thanks
$endgroup$
– Yalovetoseeit
Dec 2 '18 at 20:21


















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