Menu Close

if-7-1-1-6-1-2-5-1-3-4-

Tinku Tara 0 Comments
Pin




Question Number 185472 by liuxinnan last updated on 22/Jan/23
if ω^7 =1 (1/(ω+ω^6 ))+(1/(ω^2 +ω^5 ))+(1/(ω^3 +ω^4 ))=?
$${if}\:\omega^{\mathrm{7}} =\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\omega+\omega^{\mathrm{6}} }+\frac{\mathrm{1}}{\omega^{\mathrm{2}} +\omega^{\mathrm{5}} }+\frac{\mathrm{1}}{\omega^{\mathrm{3}} +\omega^{\mathrm{4}} }=? \\ $$
Commented by Shrinava last updated on 23/Jan/23
Give: ω^7 = 1 ⇒ ω = 1 ▲ ω^(3p+q) =ω^q ⇒ ω^2 =ω^3 =ω^4 =ω^5 =ω^6 =1 ⇒ (1/(ω+ω^6 )) + (1/(ω^2 +ω^5 )) + (1/(ω^3 +ω^4 )) = (1/2) + (1/2) + (1/2) = (3/2) = 1,5 ✓
$$\mathrm{Give}:\:\:\:\omega^{\mathrm{7}} \:=\:\mathrm{1}\:\:\Rightarrow\:\:\omega\:=\:\mathrm{1} \\ $$$$\blacktriangle\:\omega^{\mathrm{3}\boldsymbol{\mathrm{p}}+\boldsymbol{\mathrm{q}}} =\omega^{\boldsymbol{\mathrm{q}}} \\ $$$$\Rightarrow\:\omega^{\mathrm{2}} =\omega^{\mathrm{3}} =\omega^{\mathrm{4}} =\omega^{\mathrm{5}} =\omega^{\mathrm{6}} =\mathrm{1} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\omega+\omega^{\mathrm{6}} }\:+\:\frac{\mathrm{1}}{\omega^{\mathrm{2}} +\omega^{\mathrm{5}} }\:+\:\frac{\mathrm{1}}{\omega^{\mathrm{3}} +\omega^{\mathrm{4}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\:=\:\frac{\mathrm{3}}{\mathrm{2}}\:=\:\mathrm{1},\mathrm{5}\:\checkmark \\ $$
Commented by Rasheed.Sindhi last updated on 24/Jan/23
Here ω isn′t cube root of unity. (It′s seventh root of unity.) So ω^(3p+q) ≠ω^q (But ω^(7p+q) =ω^q ) Only for ω=1 your solution is correct. But ω has six other values also.
$${Here}\:\omega\:{isn}'{t}\:{cube}\:{root}\:{of}\:{unity}. \\ $$$$\left({It}'{s}\:{seventh}\:{root}\:{of}\:{unity}.\right) \\ $$$${So}\:\:\:\omega^{\mathrm{3}\boldsymbol{\mathrm{p}}+\boldsymbol{\mathrm{q}}} \neq\omega^{\boldsymbol{\mathrm{q}}} \\ $$$$\left({But}\:\omega^{\mathrm{7}\boldsymbol{\mathrm{p}}+\boldsymbol{\mathrm{q}}} =\omega^{\boldsymbol{\mathrm{q}}} \right) \\ $$$${Only}\:{for}\:\omega=\mathrm{1}\:{your}\:{solution} \\ $$$$\:{is}\:{correct}.\:{But}\:\omega\:{has}\:{six}\:{other} \\ $$$$\:{values}\:{also}. \\ $$
Answered by mr W last updated on 22/Jan/23
ω^7 =1 ⇒ω=e^((2kπi)/7) (k=0,1,2,...,6) ω+(1/ω)=2 cos ((2kπ)/7) ω^2 +(1/ω^2 )=2 cos ((4kπ)/7) ω^3 +(1/ω^3 )=2 cos ((6kπ)/7) (1/(ω+ω^6 ))+(1/(ω^2 +ω^5 ))+(1/(ω^3 +ω^4 )) =(1/(ω+(1/ω)))+(1/(ω^2 +(1/ω^2 )))+(1/(ω^3 +(1/ω^3 ))) =(1/2)((1/(cos ((2kπ)/7)))+(1/(cos ((4kπ)/7)))+(1/(cos ((6kπ)/7))))= { (((3/2), k=0)),((−2, k=1,2,...,6)) :}
$$\omega^{\mathrm{7}} =\mathrm{1} \\ $$$$\Rightarrow\omega={e}^{\frac{\mathrm{2}{k}\pi{i}}{\mathrm{7}}} \:\left({k}=\mathrm{0},\mathrm{1},\mathrm{2},…,\mathrm{6}\right) \\ $$$$\omega+\frac{\mathrm{1}}{\omega}=\mathrm{2}\:\mathrm{cos}\:\frac{\mathrm{2}{k}\pi}{\mathrm{7}} \\ $$$$\omega^{\mathrm{2}} +\frac{\mathrm{1}}{\omega^{\mathrm{2}} }=\mathrm{2}\:\mathrm{cos}\:\frac{\mathrm{4}{k}\pi}{\mathrm{7}} \\ $$$$\omega^{\mathrm{3}} +\frac{\mathrm{1}}{\omega^{\mathrm{3}} }=\mathrm{2}\:\mathrm{cos}\:\frac{\mathrm{6}{k}\pi}{\mathrm{7}} \\ $$$$\frac{\mathrm{1}}{\omega+\omega^{\mathrm{6}} }+\frac{\mathrm{1}}{\omega^{\mathrm{2}} +\omega^{\mathrm{5}} }+\frac{\mathrm{1}}{\omega^{\mathrm{3}} +\omega^{\mathrm{4}} } \\ $$$$=\frac{\mathrm{1}}{\omega+\frac{\mathrm{1}}{\omega}}+\frac{\mathrm{1}}{\omega^{\mathrm{2}} +\frac{\mathrm{1}}{\omega^{\mathrm{2}} }}+\frac{\mathrm{1}}{\omega^{\mathrm{3}} +\frac{\mathrm{1}}{\omega^{\mathrm{3}} }} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{cos}\:\frac{\mathrm{2}{k}\pi}{\mathrm{7}}}+\frac{\mathrm{1}}{\mathrm{cos}\:\frac{\mathrm{4}{k}\pi}{\mathrm{7}}}+\frac{\mathrm{1}}{\mathrm{cos}\:\frac{\mathrm{6}{k}\pi}{\mathrm{7}}}\right)=\begin{cases}{\frac{\mathrm{3}}{\mathrm{2}},\:{k}=\mathrm{0}}\\{−\mathrm{2},\:{k}=\mathrm{1},\mathrm{2},…,\mathrm{6}}\end{cases} \\ $$
Commented by liuxinnan last updated on 23/Jan/23
I cant understand the last step
$${I}\:{cant}\:{understand}\:{the}\:{last}\:{step} \\ $$
Answered by Rasheed.Sindhi last updated on 22/Jan/23
{ ((ω^7 =1)),((1+ω+ω^2 +...+ω^6 =0)) :} ▶(1/(ω+ω^6 ))+(1/(ω^2 +ω^5 ))+(1/(ω^3 +ω^4 )) =((ω^2 +ω^5 +ω+ω^6 )/((ω+ω^6 )(ω^2 +ω^5 )))+(1/(ω^3 +ω^4 )) =((−1−ω^3 −ω^4 )/(ω^3 +ω^6 +ω^8 +ω^(11) ))+(1/(ω^3 +ω^4 )) =((−1−ω^3 −ω^4 )/(ω^3 +ω^6 +ω+ω^4 ))+(1/(ω^3 +ω^4 )) =((−1−ω^3 −ω^4 )/(−1−ω^2 −ω^5 ))+(1/(ω^3 +ω^4 )) =((1+ω^3 +ω^4 )/(1+ω^2 +ω^5 ))+(1/(ω^3 +ω^4 )) =((1(ω^3 +ω^4 )+(ω^3 +ω^4 )^2 +1+ω^2 +ω^5 )/(ω^3 +ω^4 +ω^5 +ω^6 +ω^8 +ω^9 )) =((ω^3 +ω^4 +ω^6 +2ω^7 +ω^8 +1+ω^2 +ω^5 )/(ω^3 +ω^4 +ω^5 +ω^6 +ω+ω^2 )) =((ω^3 +ω^4 +ω^6 +2+ω+1+ω^2 +ω^5 )/(−1)) =−(2+0)=−2
$$\begin{cases}{\omega^{\mathrm{7}} =\mathrm{1}}\\{\mathrm{1}+\omega+\omega^{\mathrm{2}} +…+\omega^{\mathrm{6}} =\mathrm{0}}\end{cases} \\ $$$$\blacktriangleright\frac{\mathrm{1}}{\omega+\omega^{\mathrm{6}} }+\frac{\mathrm{1}}{\omega^{\mathrm{2}} +\omega^{\mathrm{5}} }+\frac{\mathrm{1}}{\omega^{\mathrm{3}} +\omega^{\mathrm{4}} } \\ $$$$=\frac{\omega^{\mathrm{2}} +\omega^{\mathrm{5}} +\omega+\omega^{\mathrm{6}} }{\left(\omega+\omega^{\mathrm{6}} \right)\left(\omega^{\mathrm{2}} +\omega^{\mathrm{5}} \right)}+\frac{\mathrm{1}}{\omega^{\mathrm{3}} +\omega^{\mathrm{4}} } \\ $$$$=\frac{−\mathrm{1}−\omega^{\mathrm{3}} −\omega^{\mathrm{4}} }{\omega^{\mathrm{3}} +\omega^{\mathrm{6}} +\omega^{\mathrm{8}} +\omega^{\mathrm{11}} }+\frac{\mathrm{1}}{\omega^{\mathrm{3}} +\omega^{\mathrm{4}} } \\ $$$$=\frac{−\mathrm{1}−\omega^{\mathrm{3}} −\omega^{\mathrm{4}} }{\omega^{\mathrm{3}} +\omega^{\mathrm{6}} +\omega+\omega^{\mathrm{4}} }+\frac{\mathrm{1}}{\omega^{\mathrm{3}} +\omega^{\mathrm{4}} } \\ $$$$=\frac{−\mathrm{1}−\omega^{\mathrm{3}} −\omega^{\mathrm{4}} }{−\mathrm{1}−\omega^{\mathrm{2}} −\omega^{\mathrm{5}} }+\frac{\mathrm{1}}{\omega^{\mathrm{3}} +\omega^{\mathrm{4}} } \\ $$$$=\frac{\mathrm{1}+\omega^{\mathrm{3}} +\omega^{\mathrm{4}} }{\mathrm{1}+\omega^{\mathrm{2}} +\omega^{\mathrm{5}} }+\frac{\mathrm{1}}{\omega^{\mathrm{3}} +\omega^{\mathrm{4}} } \\ $$$$=\frac{\mathrm{1}\left(\omega^{\mathrm{3}} +\omega^{\mathrm{4}} \right)+\left(\omega^{\mathrm{3}} +\omega^{\mathrm{4}} \right)^{\mathrm{2}} +\mathrm{1}+\omega^{\mathrm{2}} +\omega^{\mathrm{5}} }{\omega^{\mathrm{3}} +\omega^{\mathrm{4}} +\omega^{\mathrm{5}} +\omega^{\mathrm{6}} +\omega^{\mathrm{8}} +\omega^{\mathrm{9}} } \\ $$$$=\frac{\omega^{\mathrm{3}} +\omega^{\mathrm{4}} +\omega^{\mathrm{6}} +\mathrm{2}\omega^{\mathrm{7}} +\omega^{\mathrm{8}} +\mathrm{1}+\omega^{\mathrm{2}} +\omega^{\mathrm{5}} }{\omega^{\mathrm{3}} +\omega^{\mathrm{4}} +\omega^{\mathrm{5}} +\omega^{\mathrm{6}} +\omega+\omega^{\mathrm{2}} } \\ $$$$=\frac{\omega^{\mathrm{3}} +\omega^{\mathrm{4}} +\omega^{\mathrm{6}} +\mathrm{2}+\omega+\mathrm{1}+\omega^{\mathrm{2}} +\omega^{\mathrm{5}} }{−\mathrm{1}} \\ $$$$=−\left(\mathrm{2}+\mathrm{0}\right)=−\mathrm{2} \\ $$
Answered by Rasheed.Sindhi last updated on 22/Jan/23
{ ((ω^7 =1)),((1+ω+ω^2 +...+ω^6 =0)) :} ▶(1/(ω+ω^6 ))+(1/(ω^2 +ω^5 ))+(1/(ω^3 +ω^4 )) =(ω/(ω^2 +ω^7 ))+(ω^2 /(ω^4 +ω^7 ))+(ω^3 /(ω^6 +ω^7 )) =(ω/(1+ω^2 ))+(ω^2 /(1+ω^4 ))+(ω^3 /(1+ω^6 )) =((1+ω^4 +1+ω^2 )/(1+ω^4 +ω^2 +ω^6 ))−1+1+(ω^3 /(1+ω^6 )) =((1−ω^6 )/(1+ω^4 +ω^2 +ω^6 ))+1+(ω^3 /(1+ω^6 )) =((1−ω^6 )/(−ω−ω^3 −ω^5 ))+(ω^3 /(1+ω^6 ))+1 =((ω^6 −1)/(ω+ω^3 +ω^5 ))+(ω^3 /(1+ω^6 ))+1 =((ω^(12) −1+ω^4 +ω^6 +ω^8 )/(ω+ω^3 +ω^5 +ω^7 +ω^9 +ω^(11) ))+1 =((ω^5 −1+ω^4 +ω^6 +ω)/(ω+ω^3 +ω^5 +1+ω^2 +ω^4 ))+1 =((−1−1−ω^2 −ω^3 )/(−ω^6 ))+1 =((2+ω^2 +ω^3 )/ω^6 )+1 =2ω+ω^3 +ω^4 +1
$$\begin{cases}{\omega^{\mathrm{7}} =\mathrm{1}}\\{\mathrm{1}+\omega+\omega^{\mathrm{2}} +…+\omega^{\mathrm{6}} =\mathrm{0}}\end{cases} \\ $$$$\blacktriangleright\frac{\mathrm{1}}{\omega+\omega^{\mathrm{6}} }+\frac{\mathrm{1}}{\omega^{\mathrm{2}} +\omega^{\mathrm{5}} }+\frac{\mathrm{1}}{\omega^{\mathrm{3}} +\omega^{\mathrm{4}} } \\ $$$$=\frac{\omega}{\omega^{\mathrm{2}} +\omega^{\mathrm{7}} }+\frac{\omega^{\mathrm{2}} }{\omega^{\mathrm{4}} +\omega^{\mathrm{7}} }+\frac{\omega^{\mathrm{3}} }{\omega^{\mathrm{6}} +\omega^{\mathrm{7}} } \\ $$$$=\frac{\omega}{\mathrm{1}+\omega^{\mathrm{2}} }+\frac{\omega^{\mathrm{2}} }{\mathrm{1}+\omega^{\mathrm{4}} }+\frac{\omega^{\mathrm{3}} }{\mathrm{1}+\omega^{\mathrm{6}} } \\ $$$$=\frac{\mathrm{1}+\omega^{\mathrm{4}} +\mathrm{1}+\omega^{\mathrm{2}} }{\mathrm{1}+\omega^{\mathrm{4}} +\omega^{\mathrm{2}} +\omega^{\mathrm{6}} }−\mathrm{1}+\mathrm{1}+\frac{\omega^{\mathrm{3}} }{\mathrm{1}+\omega^{\mathrm{6}} } \\ $$$$=\frac{\mathrm{1}−\omega^{\mathrm{6}} }{\mathrm{1}+\omega^{\mathrm{4}} +\omega^{\mathrm{2}} +\omega^{\mathrm{6}} }+\mathrm{1}+\frac{\omega^{\mathrm{3}} }{\mathrm{1}+\omega^{\mathrm{6}} } \\ $$$$=\frac{\mathrm{1}−\omega^{\mathrm{6}} }{−\omega−\omega^{\mathrm{3}} −\omega^{\mathrm{5}} }+\frac{\omega^{\mathrm{3}} }{\mathrm{1}+\omega^{\mathrm{6}} }+\mathrm{1} \\ $$$$=\frac{\omega^{\mathrm{6}} −\mathrm{1}}{\omega+\omega^{\mathrm{3}} +\omega^{\mathrm{5}} }+\frac{\omega^{\mathrm{3}} }{\mathrm{1}+\omega^{\mathrm{6}} }+\mathrm{1} \\ $$$$=\frac{\omega^{\mathrm{12}} −\mathrm{1}+\omega^{\mathrm{4}} +\omega^{\mathrm{6}} +\omega^{\mathrm{8}} }{\omega+\omega^{\mathrm{3}} +\omega^{\mathrm{5}} +\omega^{\mathrm{7}} +\omega^{\mathrm{9}} +\omega^{\mathrm{11}} }+\mathrm{1} \\ $$$$=\frac{\omega^{\mathrm{5}} −\mathrm{1}+\omega^{\mathrm{4}} +\omega^{\mathrm{6}} +\omega}{\omega+\omega^{\mathrm{3}} +\omega^{\mathrm{5}} +\mathrm{1}+\omega^{\mathrm{2}} +\omega^{\mathrm{4}} }+\mathrm{1} \\ $$$$=\frac{−\mathrm{1}−\mathrm{1}−\omega^{\mathrm{2}} −\omega^{\mathrm{3}} }{−\omega^{\mathrm{6}} }+\mathrm{1} \\ $$$$=\frac{\mathrm{2}+\omega^{\mathrm{2}} +\omega^{\mathrm{3}} }{\omega^{\mathrm{6}} }+\mathrm{1} \\ $$$$=\mathrm{2}\omega+\omega^{\mathrm{3}} +\omega^{\mathrm{4}} +\mathrm{1} \\ $$
Answered by Rasheed.Sindhi last updated on 22/Jan/23
if ω^7 =1 { ((ω^7 =1)),((1+ω+ω^2 +...ω^6 =0)) :} ▶(1/(ω+ω^6 ))+(1/(ω^2 +ω^5 ))+(1/(ω^3 +ω^4 )) =(1/(ω+ω^6 ))∙((ω−ω^6 )/(ω−ω^6 ))+(1/(ω^2 +ω^5 ))+(1/(ω^3 +ω^4 )) =((ω−ω^6 )/(ω^2 −ω^(12) ))+(1/(ω^2 +ω^5 ))+(1/(ω^3 +ω^4 )) =((ω−ω^6 )/(ω^2 −ω^5 ))+(1/(ω^2 +ω^5 ))+(1/(ω^3 +ω^4 )) =((ω^3 +ω^6 −ω^8 −ω^(11) )/(ω^4 −ω^(10) ))+(1/(ω^3 +ω^4 )) =((ω^3 +ω^6 −ω−ω^4 )/(ω^4 −ω^3 ))+(1/(ω^3 +ω^4 )) =((ω^6 +ω^9 −ω^4 −ω^7 +ω^7 +ω^(10) −ω^5 −ω^8 )/(ω^8 −ω^6 )) =((ω^6 +ω^2 −ω^4 −ω^7 +ω^7 +ω^3 −ω^5 −ω)/(ω−ω^6 )) =((ω^6 +ω^3 +ω^2 −ω^4 −ω^5 −ω)/(ω−ω^6 ))+1−1 =((ω^3 +ω^2 −(ω^4 +ω^5 ))/(ω−ω^6 ))−1 =((ω^3 +ω^2 −(−1−ω−ω^2 −ω^3 −ω^6 ))/(ω−ω^6 ))−1 =((ω^3 +ω^2 +1+ω+ω^2 +ω^3 +ω^6 )/(ω−ω^6 ))−1 =((2ω^3 +2ω^2 +1+2ω^6 )/(ω−ω^6 ))
$${if}\:\omega^{\mathrm{7}} =\mathrm{1} \\ $$$$\begin{cases}{\omega^{\mathrm{7}} =\mathrm{1}}\\{\mathrm{1}+\omega+\omega^{\mathrm{2}} +…\omega^{\mathrm{6}} =\mathrm{0}}\end{cases} \\ $$$$\blacktriangleright\frac{\mathrm{1}}{\omega+\omega^{\mathrm{6}} }+\frac{\mathrm{1}}{\omega^{\mathrm{2}} +\omega^{\mathrm{5}} }+\frac{\mathrm{1}}{\omega^{\mathrm{3}} +\omega^{\mathrm{4}} } \\ $$$$=\frac{\mathrm{1}}{\omega+\omega^{\mathrm{6}} }\centerdot\frac{\omega−\omega^{\mathrm{6}} }{\omega−\omega^{\mathrm{6}} }+\frac{\mathrm{1}}{\omega^{\mathrm{2}} +\omega^{\mathrm{5}} }+\frac{\mathrm{1}}{\omega^{\mathrm{3}} +\omega^{\mathrm{4}} } \\ $$$$=\frac{\omega−\omega^{\mathrm{6}} }{\omega^{\mathrm{2}} −\omega^{\mathrm{12}} }+\frac{\mathrm{1}}{\omega^{\mathrm{2}} +\omega^{\mathrm{5}} }+\frac{\mathrm{1}}{\omega^{\mathrm{3}} +\omega^{\mathrm{4}} } \\ $$$$=\frac{\omega−\omega^{\mathrm{6}} }{\omega^{\mathrm{2}} −\omega^{\mathrm{5}} }+\frac{\mathrm{1}}{\omega^{\mathrm{2}} +\omega^{\mathrm{5}} }+\frac{\mathrm{1}}{\omega^{\mathrm{3}} +\omega^{\mathrm{4}} } \\ $$$$=\frac{\omega^{\mathrm{3}} +\omega^{\mathrm{6}} −\omega^{\mathrm{8}} −\omega^{\mathrm{11}} }{\omega^{\mathrm{4}} −\omega^{\mathrm{10}} }+\frac{\mathrm{1}}{\omega^{\mathrm{3}} +\omega^{\mathrm{4}} } \\ $$$$=\frac{\omega^{\mathrm{3}} +\omega^{\mathrm{6}} −\omega−\omega^{\mathrm{4}} }{\omega^{\mathrm{4}} −\omega^{\mathrm{3}} }+\frac{\mathrm{1}}{\omega^{\mathrm{3}} +\omega^{\mathrm{4}} } \\ $$$$=\frac{\omega^{\mathrm{6}} +\omega^{\mathrm{9}} −\omega^{\mathrm{4}} −\omega^{\mathrm{7}} +\omega^{\mathrm{7}} +\omega^{\mathrm{10}} −\omega^{\mathrm{5}} −\omega^{\mathrm{8}} }{\omega^{\mathrm{8}} −\omega^{\mathrm{6}} } \\ $$$$=\frac{\omega^{\mathrm{6}} +\omega^{\mathrm{2}} −\omega^{\mathrm{4}} −\cancel{\omega^{\mathrm{7}} }+\cancel{\omega^{\mathrm{7}} }+\omega^{\mathrm{3}} −\omega^{\mathrm{5}} −\omega}{\omega−\omega^{\mathrm{6}} } \\ $$$$=\frac{\omega^{\mathrm{6}} +\omega^{\mathrm{3}} +\omega^{\mathrm{2}} −\omega^{\mathrm{4}} −\omega^{\mathrm{5}} −\omega}{\omega−\omega^{\mathrm{6}} }+\mathrm{1}−\mathrm{1} \\ $$$$=\frac{\omega^{\mathrm{3}} +\omega^{\mathrm{2}} −\left(\omega^{\mathrm{4}} +\omega^{\mathrm{5}} \right)}{\omega−\omega^{\mathrm{6}} }−\mathrm{1} \\ $$$$=\frac{\omega^{\mathrm{3}} +\omega^{\mathrm{2}} −\left(−\mathrm{1}−\omega−\omega^{\mathrm{2}} −\omega^{\mathrm{3}} −\omega^{\mathrm{6}} \right)}{\omega−\omega^{\mathrm{6}} }−\mathrm{1} \\ $$$$=\frac{\omega^{\mathrm{3}} +\omega^{\mathrm{2}} +\mathrm{1}+\omega+\omega^{\mathrm{2}} +\omega^{\mathrm{3}} +\omega^{\mathrm{6}} }{\omega−\omega^{\mathrm{6}} }−\mathrm{1} \\ $$$$=\frac{\mathrm{2}\omega^{\mathrm{3}} +\mathrm{2}\omega^{\mathrm{2}} +\mathrm{1}+\mathrm{2}\omega^{\mathrm{6}} }{\omega−\omega^{\mathrm{6}} } \\ $$
Commented by liuxinnan last updated on 23/Jan/23
thanks
$${thanks} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *