Question Number 2187 by 123456 last updated on 07/Nov/15 $${f}:\mathbb{N}\rightarrow\mathbb{N} \\ $$$${f}\left({n}+{m}\right)={f}\left({n}\right)+\alpha{m} \\ $$$${f}\left(\mathrm{0}\right)={k} \\ $$$${f}\left({n}\right)=? \\ $$ Answered by prakash jain last updated on…
Question Number 67672 by Abdo msup. last updated on 30/Aug/19 $${decompose}\:{the}\:{folowing}\:\:{fraction}\:{at}\:{R}\left({x}\right) \\ $$$$\left.\mathrm{1}\right){F}\left({x}\right)=\frac{{x}^{\mathrm{3}} }{\mathrm{1}−{x}^{\mathrm{6}} } \\ $$$$\left.\mathrm{2}\right)\:{G}\left({x}\right)\:=\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{3}} \left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$ Commented by…
Question Number 133168 by bemath last updated on 19/Feb/21 Answered by floor(10²Eta[1]) last updated on 19/Feb/21 $$\mathrm{x}\rightarrow\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}} \\ $$$$\left(\mathrm{I}\right):\:\mathrm{f}\left(\frac{\mathrm{x}}{\mathrm{x}−\mathrm{1}}\right)+\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)=\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}} \\ $$$$\mathrm{but}\:\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)=\mathrm{x}−\mathrm{f}\left(\mathrm{1}−\mathrm{x}\right),\:\mathrm{so} \\ $$$$\mathrm{f}\left(\frac{\mathrm{x}}{\mathrm{x}−\mathrm{1}}\right)+\mathrm{x}−\mathrm{f}\left(\mathrm{1}−\mathrm{x}\right)=\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}} \\ $$$$\left(\mathrm{II}\right):\:\mathrm{f}\left(\frac{\mathrm{x}}{\mathrm{x}−\mathrm{1}}\right)−\mathrm{f}\left(\mathrm{1}−\mathrm{x}\right)=\frac{−\mathrm{x}^{\mathrm{2}}…
Question Number 133127 by abdomsup last updated on 19/Feb/21 $${find}\:{lim}_{{n}\rightarrow\infty} \int_{\mathrm{0}} ^{{n}} \left(\mathrm{1}−\frac{{x}}{{n}}\right)^{{n}} {lnx}\:{dx} \\ $$ Answered by Dwaipayan Shikari last updated on 19/Feb/21 $$\underset{{n}\rightarrow\infty}…
Question Number 133122 by abdomsup last updated on 19/Feb/21 $${let}\:{V}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:\frac{\mathrm{1}}{\:\sqrt{{n}^{\mathrm{2}} +\mathrm{1}}} \\ $$$${find}\:{a}\:{ewivalent}\:{of}\:{V}_{{n}} \left({n}\smile\infty\right) \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 67559 by Abdo msup. last updated on 28/Aug/19 $${calculate}\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}^{\mathrm{2}\:} +\mathrm{1}}\:\:{and}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} \:+\mathrm{1}} \\ $$ Commented by ~ À ®…
Question Number 67540 by mathmax by abdo last updated on 28/Aug/19 $${prove}\:{that}\:\:\:\mid\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}+{it}\right)\mid\:=\sqrt{\frac{\mathrm{2}\pi}{{e}^{\pi{t}} \:+{e}^{−\pi{t}} }} \\ $$$${and}\:\mid\Gamma\left(\mathrm{1}+{it}\right)\mid\:=\sqrt{\frac{\mathrm{2}\pi{t}}{{e}^{\pi{t}} −{e}^{−\pi{t}} }} \\ $$ Commented by ~ À ®…
Question Number 67538 by mathmax by abdo last updated on 28/Aug/19 $${prove}\:{that}\:\frac{\Gamma^{'} \left({z}\right)}{\Gamma\left({z}\right)}\:=−\gamma−\frac{\mathrm{1}}{{z}}\:−\sum_{{n}=\mathrm{1}} ^{\infty} \left(\frac{\mathrm{1}}{{z}+{n}}−\frac{\mathrm{1}}{{n}}\right) \\ $$ Commented by ~ À ® @ 237 ~…
Question Number 67537 by mathmax by abdo last updated on 28/Aug/19 $${prove}\:{that}\:\frac{\mathrm{1}}{\Gamma\left({z}\right)}\:={z}\:{e}^{\gamma{z}} \:\prod_{{n}=\mathrm{1}} ^{\infty} \left(\mathrm{1}+\frac{{z}}{{n}}\right){e}^{−\frac{{z}}{{n}}} \\ $$ Commented by ~ À ® @ 237 ~…
Question Number 67534 by mathmax by abdo last updated on 28/Aug/19 $${find}\:{the}\:{value}\:{of}\:\:\prod_{{n}=\mathrm{2}} ^{\infty} \:\frac{{n}^{\mathrm{3}} −\mathrm{1}}{{n}^{\mathrm{3}} \:+\mathrm{1}} \\ $$$${and}\:\prod_{{n}=\mathrm{1}} ^{\infty} \left(\mathrm{1}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right) \\ $$ Commented by…