if $x^5=1$ with $x\neq 1$ then find value of $\frac{x}{1+x^2}+\frac{x^2}{1+x^4}+\frac{x^3}{1+x}+\frac{x^4}{1+x^3}$

Question

if $x^5=1$ with $x\neq 1$ then find value of $$\frac{x}{1+x^2}+\frac{x^2}{1+x^4}+\frac{x^3}{1+x}+\frac{x^4}{1+x^3}$$

So my first observation was x is a non real fifth root of unity. Also $$x^5-1=(x-1)(1+x+x^2+x^3+x^4)=0$$

Thus $$1+x+x^2+x^3+x^4=0$$ I tried using this condition to simplify the above expression but nothing interesting simplified. Please note i am looking for hints rather than complete solutions.

EDIT:I came to know its a duplicate ,but i feel that the answers given below are different from those in the original.

@Äres it does but i dont want it to be flagged as duplicate as different answers from those you gave have been proposed below — Albus Dumbledore, Sep 29 '20 at 11:22
@AlbusDumbledore The question is duplicate. I think you can ask for merging...I have had a similar experience here — Anindya Prithvi, Sep 29 '20 at 12:59

score 11 · Accepted Answer · answered Sep 29 '20 at 10:22

$$\frac{x}{1+x^2}+\frac{x^2}{1+x^4}+\frac{x^3}{1+x}+\frac{x^4}{1+x^3} =$$ $$=\frac{x}{1+x^2}\cdot\frac{x^4}{x^4}+\frac{x^2}{1+x^4}\cdot\frac{x^3}{x^3}+\frac{x^3}{1+x}\cdot \frac{x^2}{x^2}+\frac{x^4}{1+x^3}\cdot\frac{x}{x}= $$ (remember that $x^5=1$, so $x^6=x$ and $x^7=x^2$) $$=\frac{1}{x^4+x}+\frac{1}{x^3+x^2}+\frac{1}{x^2+x^3}+\frac{1}{x+x^4}= 2\left(\frac{1}{x+x^4}+\frac{1}{x^2+x^3}\right)=$$ $$= 2\left(\frac{x^2+x^3+x+x^4}{(x+x^4)(x^2+x^3)} \right) = 2\left(\frac{x+x^2+x^3+x^4}{x^3+x^4+x^6+x^7} \right) = 2\left(\frac{x+x^2+x^3+x^4}{x^3+x^4+x+x^2} \right) =2. $$

score 9 · Answer 2 · answered Sep 29 '20 at 10:53

9

Note that $$ \frac1{1+x^n}=\frac12\frac{1+x^{5n}}{1+x^n}=\frac{1-x^n+x^{2n}-x^{3n}+x^{4n}}2\tag1 $$ Applying $(1)$ gives $$ \begin{align} \frac{x}{1+x^2} &=\frac{x-x^3+x^5-x^7+x^9}2\\ &=\frac{x-x^3+1-x^2+x^4}2\tag2 \end{align} $$ $$ \begin{align} \frac{x^2}{1+x^4} &=\frac{x^2-x^6+x^{10}-x^{14}+x^{18}}2\\ &=\frac{x^2-x+1-x^4+x^3}2\tag3 \end{align} $$ $$ \begin{align} \frac{x^3}{1+x} &=\frac{x^3-x^4+x^5-x^6+x^7}2\\ &=\frac{x^3-x^4+1-x+x^2}2\tag4 \end{align} $$ $$ \begin{align} \frac{x^4}{1+x^3} &=\frac{x^4-x^7+x^{10}-x^{13}+x^{16}}2\\ &=\frac{x^4-x^2+1-x^3+x}2\tag5 \end{align} $$ Each power of $x$ appears twice positive and twice negative, except for $1$ which is always positive. Therefore, $$ \begin{align} \frac{x}{1+x^2}+\frac{x^2}{1+x^4}+\frac{x^3}{1+x}+\frac{x^4}{1+x^3} &=\frac{1+1+1+1}2\\ &=2\tag6 \end{align} $$

answered Sep 29 '20 at 10:53

robjohn

345,667

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A very unique answer ,Thank you! – Albus Dumbledore Sep 29 '20 at 11:22
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Thanks. A strategy I try in situations like this, is to write things using the basis $\left{1,x,x^2,x^3,x^4\right}$ (or whatever applies to the situation). – robjohn Sep 29 '20 at 11:28
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Really lovely solution, thank you. – A-Level Student Sep 29 '20 at 12:25
If $x^5=1$ and $x\ne1$ and $5\nmid n$, $\frac1{1-x^n}=\frac{4+3x^n+2x^{2n}+x^{3n}}5$ – robjohn Sep 29 '20 at 19:06

score 1 · Answer 3 · edited Sep 29 '20 at 16:24

1

Answer :

((x / (1+x² ))+((x³/ (1+x))=((x+x² +x³ +x⁵) / ((1+x)(1+x²) ))=((x+x² +x³ +x⁵) / (x+x² +x³ +1))=1

Because x^5=1

$((x² / (1+x⁴))+((x⁴/ (1+x³ )) = ((x² +x⁵+x⁴+x⁸) / (1+x³ +x⁴+x^7)) =((x² +1+x⁴+x³) / (1+x³+x⁴+x²)) =1$

Because $x^5 =1$

$\frac{x}{1+x^2 }+\frac{x^3 }{1+x} +\frac{x^2 }{1 +x^4}+\frac{x^4}{1+x^3 }=\frac{x+x^2 +x^3 +x^5}{(1+x)(1+x^2) }+\frac{x^2 +x^5+x^4+x^8}{(1+x^4)(1+x^3) }=\frac{x+x^2 +x^3 +x^5}{x+x^2 +x^3 +1}+\frac{x^2 +1+x^4+x^3 }{1+x^3 +x^4+x^2} =2$ Because $x^5 =1\Rightarrow x^7=x^2$ and $x ^8 = x^3$

edited Sep 29 '20 at 16:24

Albus Dumbledore

11,860

answered Sep 29 '20 at 14:32

Anas anas

21
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Thank you for trying but do you mind using mathjax for better formatting,thanks! – Albus Dumbledore Sep 29 '20 at 14:47
My answer is very simple than other answers, but I can't us mathjax – Anas anas Sep 29 '20 at 15:53
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https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Albus Dumbledore Sep 29 '20 at 15:54
See my answer now – Anas anas Sep 29 '20 at 16:21
yes it is nice!Thank you! But don't forget to enclose your expressions with '$',.I have edited please see.Thnks for the effort – Albus Dumbledore Sep 29 '20 at 16:26
Okey, thank you – Anas anas Sep 29 '20 at 16:29

if $x^5=1$ with $x\neq 1$ then find value of $\frac{x}{1+x^2}+\frac{x^2}{1+x^4}+\frac{x^3}{1+x}+\frac{x^4}{1+x^3}$

3 Answers3