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Question Number 203742 by Calculusboy last updated on 27/Jan/24

Answered by mr W last updated on 27/Jan/24

x^(2024) +x^(2024) −2024×(1/4)x^(2023) +...=0 2x^(2024) −506x^(2023) +...=0 sum of roots =((506)/2)=253 ✓ solution: ((1/(4x))−1)^(2024) =−1=e^((2k+1)πi) (1/(4x))−1=e^(((2k+1)πi)/(2024)) ⇒x=(1/(4[1+e^(((2k+1)πi)/(2024)) ])) with k=0,1,...,2023

$${x}^{\mathrm{2024}} +{x}^{\mathrm{2024}} −\mathrm{2024}×\frac{\mathrm{1}}{\mathrm{4}}{x}^{\mathrm{2023}} +...=\mathrm{0} \\ $$$$\mathrm{2}{x}^{\mathrm{2024}} −\mathrm{506}{x}^{\mathrm{2023}} +...=\mathrm{0} \\ $$$${sum}\:{of}\:{roots}\:=\frac{\mathrm{506}}{\mathrm{2}}=\mathrm{253}\:\checkmark \\ $$$$ \\ $$$${solution}: \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{4}{x}}−\mathrm{1}\right)^{\mathrm{2024}} =−\mathrm{1}={e}^{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi{i}} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}{x}}−\mathrm{1}={e}^{\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi{i}}{\mathrm{2024}}} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{4}\left[\mathrm{1}+{e}^{\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi{i}}{\mathrm{2024}}} \right]}\:\:{with}\:{k}=\mathrm{0},\mathrm{1},...,\mathrm{2023} \\ $$

Commented by Calculusboy last updated on 27/Jan/24

thanks sir

$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$

Answered by a.lgnaoui last updated on 27/Jan/24

x^(2024) +(((1−4x)^(2024) )/4^(2024) )=0 x^(2024) =−((1/4)−x)^(2024) ( ((4x)/(1−4x)))^(2024) =−1=(i^2 ) ( ((4x)/(1−4x)))=(i)^(1/(1012)) ⇒ (1/(4x))=(1/i^(1/(1012)) )+1=((1+i^(1/(1012)) )/i^(1/(1012)) ) x =(e^((𝛑/(2024))i) /(4(1+e^((𝛑/(2024))i) )))

$$\boldsymbol{\mathrm{x}}^{\mathrm{2024}} +\frac{\left(\mathrm{1}−\mathrm{4}\boldsymbol{\mathrm{x}}\right)^{\mathrm{2024}} }{\mathrm{4}^{\mathrm{2024}} }=\mathrm{0} \\ $$$$\:\:\:\boldsymbol{\mathrm{x}}^{\mathrm{2024}} =−\left(\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{x}\right)^{\mathrm{2024}} \\ $$$$\:\left(\:\:\frac{\mathrm{4}\boldsymbol{\mathrm{x}}}{\mathrm{1}−\mathrm{4}\boldsymbol{\mathrm{x}}}\right)^{\mathrm{2024}} =−\mathrm{1}=\left(\boldsymbol{\mathrm{i}}^{\mathrm{2}} \right) \\ $$$$\:\left(\:\frac{\mathrm{4}\boldsymbol{\mathrm{x}}}{\mathrm{1}−\mathrm{4}\boldsymbol{\mathrm{x}}}\right)=\left(\boldsymbol{\mathrm{i}}\right)^{\frac{\mathrm{1}}{\mathrm{1012}}} \\ $$$$\Rightarrow\:\:\:\frac{\mathrm{1}}{\mathrm{4x}}=\frac{\mathrm{1}}{\mathrm{i}^{\frac{\mathrm{1}}{\mathrm{1012}}} }+\mathrm{1}=\frac{\mathrm{1}+\mathrm{i}^{\frac{\mathrm{1}}{\mathrm{1012}}} }{\mathrm{i}^{\frac{\mathrm{1}}{\mathrm{1012}}} } \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\boldsymbol{\mathrm{x}}\:\:\:\:\:=\frac{\boldsymbol{\mathrm{e}}^{\frac{\boldsymbol{\pi}}{\mathrm{2024}}\boldsymbol{\mathrm{i}}} }{\mathrm{4}\left(\mathrm{1}+\mathrm{e}^{\frac{\boldsymbol{\pi}}{\mathrm{2024}}\boldsymbol{\mathrm{i}}} \right)} \\ $$$$\:\:\:\:\:\:\:\: \\ $$

Commented by mr W last updated on 27/Jan/24

there are not only one, but totally 2024 complex roots!

$${there}\:{are}\:{not}\:{only}\:{one},\:{but}\:{totally} \\ $$$$\mathrm{2024}\:{complex}\:{roots}! \\ $$

Commented by Calculusboy last updated on 27/Jan/24

thanks sir

$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$

Commented by a.lgnaoui last updated on 27/Jan/24

yes thanks x=(e^((π(2k+1)i)/(2024)) /((1+e^((2k+1)(π/(2024))) ))) (k=0 to 2023)

$$\mathrm{yes}\:\mathrm{thanks}\: \\ $$$$ \\ $$$$\:\mathrm{x}=\frac{\mathrm{e}^{\frac{\pi\left(\mathrm{2k}+\mathrm{1}\right)\mathrm{i}}{\mathrm{2024}}} }{\left(\mathrm{1}+\mathrm{e}^{\left(\mathrm{2k}+\mathrm{1}\right)\frac{\pi}{\mathrm{2024}}} \right)}\:\:\:\:\:\:\:\:\left(\mathrm{k}=\mathrm{0}\:\mathrm{to}\:\mathrm{2023}\right) \\ $$

Answered by MathematicalUser2357 last updated on 28/Jan/24

x=(e^((π(2k+1)i)/(2024)) /(1+e^(((2k+1)π)/(2024)) ))∧0≤k<2023∧∀k∈N

$${x}=\frac{{e}^{\frac{\pi\left(\mathrm{2}{k}+\mathrm{1}\right){i}}{\mathrm{2024}}} }{\mathrm{1}+{e}^{\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{2024}}} }\wedge\mathrm{0}\leq{k}<\mathrm{2023}\wedge\forall{k}\in\mathbb{N} \\ $$

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