0-lt-x-lt-1-1-1-x-1-2x-1-x-2-4x-3-1-x-4-8x-7-1-x-8-16x-15-1-x-16-evaluate-the-previous-summation-

Question Number 195651 by York12 last updated on 06/Aug/23

0<x<1 (1/(1+x^1 ))+((2x)/(1+x^2 ))+((4x^3 )/(1+x^4 ))+((8x^7 )/(1+x^8 ))+((16x^(15) )/(1+x^(16) ))+....+∞ evaluate the previous summation

$$\mathrm{0}<{x}<\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{1}} }+\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }+\frac{\mathrm{4}{x}^{\mathrm{3}} }{\mathrm{1}+{x}^{\mathrm{4}} }+\frac{\mathrm{8}{x}^{\mathrm{7}} }{\mathrm{1}+{x}^{\mathrm{8}} }+\frac{\mathrm{16}{x}^{\mathrm{15}} }{\mathrm{1}+{x}^{\mathrm{16}} }+….+\infty \\ $$$${evaluate}\:{the}\:{previous}\:{summation} \\ $$

Answered by witcher3 last updated on 06/Aug/23

Σ_(n≥0) ((2^n x^(2^n −1) )/(1+x^2^n ))=f(x) ∫_0 ^t f(x)dx=Σ_(n≥0) ln(1+t^2^n )=ln(Π_(n≥0) (1+t^2^n )) 1+x=((1−x^2 )/(1−x)) =ln(Π_(n≥0) ((1−t^2^(n+1) )/(1−t^2^n )))=lim_(N→∞) ln(Π_0 ^N ((1−t^2^(n+1) )/(1−t^2^n ))) =lim_(N→∞) ln(((1−t^2^(N+1) )/(1−t)))=−ln(1−t)=∫_0 ^t f(x)dx f(x)=(1/(1−x))

$$\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{2}^{\mathrm{n}} \mathrm{x}^{\mathrm{2}^{\mathrm{n}} −\mathrm{1}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}^{\mathrm{n}} } }=\mathrm{f}\left(\mathrm{x}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{t}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\mathrm{ln}\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}^{\mathrm{n}} } \right)=\mathrm{ln}\left(\underset{\mathrm{n}\geqslant\mathrm{0}} {\prod}\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}^{\mathrm{n}} } \right)\right) \\ $$$$\mathrm{1}+\mathrm{x}=\frac{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }{\mathrm{1}−\mathrm{x}} \\ $$$$=\mathrm{ln}\left(\underset{\mathrm{n}\geqslant\mathrm{0}} {\prod}\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}^{\mathrm{n}+\mathrm{1}} } }{\mathrm{1}−\mathrm{t}^{\mathrm{2}^{\mathrm{n}} } }\right)=\underset{\mathrm{N}\rightarrow\infty} {\mathrm{lim}ln}\left(\underset{\mathrm{0}} {\overset{\mathrm{N}} {\prod}}\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}^{\mathrm{n}+\mathrm{1}} } }{\mathrm{1}−\mathrm{t}^{\mathrm{2}^{\mathrm{n}} } }\right) \\ $$$$=\underset{\mathrm{N}\rightarrow\infty} {\mathrm{lim}ln}\left(\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}^{\mathrm{N}+\mathrm{1}} } }{\mathrm{1}−\mathrm{t}}\right)=−\mathrm{ln}\left(\mathrm{1}−\mathrm{t}\right)=\int_{\mathrm{0}} ^{\mathrm{t}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}} \\ $$

Commented by York12 last updated on 07/Aug/23

$${should}\:{I}\:{get}\:{through}\:{calculus} \\ $$$${or}\:{I}\:{can}\:{skip}\:{it}\:{and}\:{study}\:{real}\:{analysis} \\ $$

Commented by witcher3 last updated on 07/Aug/23

in Maths You can′t skip lesson all are usufull Geometrie related algebra withe analysis withe number theory geometrie &algebra Geometric−algebra number theorie +analysis=modular forms algebra and analysis=K Theory k theory is not used so much one if the hardest Topic in Maths

$$\mathrm{in}\:\mathrm{Maths}\:\mathrm{You}\:\mathrm{can}'\mathrm{t}\:\mathrm{skip}\:\mathrm{lesson}\:\mathrm{all}\:\mathrm{are}\:\mathrm{usufull} \\ $$$$\mathrm{Geometrie}\:\mathrm{related}\:\mathrm{algebra}\:\mathrm{withe}\:\mathrm{analysis} \\ $$$$\mathrm{withe}\:\mathrm{number}\:\mathrm{theory} \\ $$$$\mathrm{geometrie}\:\&\mathrm{algebra}\:\mathrm{Geometric}−\mathrm{algebra} \\ $$$$\mathrm{number}\:\mathrm{theorie}\:+\mathrm{analysis}=\mathrm{modular}\:\mathrm{forms} \\ $$$$\mathrm{algebra}\:\mathrm{and}\:\mathrm{analysis}=\mathrm{K}\:\mathrm{Theory} \\ $$$$\mathrm{k}\:\mathrm{theory}\:\mathrm{is}\:\mathrm{not}\:\mathrm{used}\:\mathrm{so}\:\mathrm{much}\:\mathrm{one}\:\mathrm{if}\:\mathrm{the}\:\mathrm{hardest}\:\mathrm{Topic} \\ $$$$\mathrm{in}\:\mathrm{Maths} \\ $$

Commented by York12 last updated on 08/Aug/23