Question Number 10563 by FilupS last updated on 18/Feb/17 $$\mathrm{can}\:\mathrm{someone}\:\mathrm{explain}\:\mathrm{to}\:\mathrm{me} \\ $$$$\mathrm{big}\:\mathrm{K}\:\mathrm{notation}?\:\left(\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{the}\:\mathrm{name}\right) \\ $$$$\mathrm{It}\:\mathrm{is}\:\mathrm{related}\:\mathrm{to}\:\mathrm{continuous}\:\mathrm{fractions}. \\ $$$$\mathrm{e}.\mathrm{g}.\:\:\:{x}={b}_{\mathrm{0}} +\underset{{i}=\mathrm{1}} {\overset{\infty} {\boldsymbol{\mathrm{K}}}}\frac{{a}_{{i}} }{{b}_{{i}} } \\ $$$$\: \\ $$$${e}^{{x}}…
Question Number 141632 by Willson last updated on 21/May/21 $$\mathrm{Let}\:{f}\left({x}\right)=\frac{{sin}\left({x}\right)}{{x}}\:,\:\mathrm{prove}\:\mathrm{that}\:: \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\:{f}\left({n}\pi+\alpha\right)+{f}\left({n}\pi−\alpha\right)\:\right]=\:\mathrm{1}+{f}\left(\alpha\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 10562 by FilupS last updated on 18/Feb/17 $$\boldsymbol{{x}}=\begin{bmatrix}{{x}_{\mathrm{1}} }\\{{x}_{\mathrm{2}} }\\{\:\vdots}\\{{x}_{{n}} }\end{bmatrix}\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{y}}=\begin{bmatrix}{{y}_{\mathrm{1}} }\\{{y}_{\mathrm{2}} }\\{\:\vdots}\\{{y}_{{n}} }\end{bmatrix}\:\:\:\:\:\:\:\:\boldsymbol{{x}},\:\boldsymbol{{y}}\:\in\:\mathbb{R}^{{n}} \\ $$$$\: \\ $$$$\mathrm{1}.\:\mathrm{Prove}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{vector}\:\boldsymbol{{x}},\:\mathrm{denoted}\:\mid\mid\boldsymbol{{x}}\mid\mid, \\ $$$$\:\:\:\:\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\sqrt{{x}_{\mathrm{1}} ^{\mathrm{2}} +{x}_{\mathrm{2}} ^{\mathrm{2}}…
Question Number 141635 by learner001 last updated on 21/May/21 $${solve}\:{the}\:{differential}\:{equation}\:\left({PDE}\right), \\ $$$${z}\left(\frac{\partial{z}}{\partial{x}}−\frac{\partial{z}}{\partial{y}}\right)={z}^{\mathrm{2}} +\left({x}+{y}\right)^{\mathrm{2}} . \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 141628 by mathdanisur last updated on 21/May/21 Commented by MJS_new last updated on 21/May/21 $$\mathrm{what}\:\mathrm{are}\:{r},\:{r}_{{a}} ,\:…? \\ $$$${rr}_{{a}} \:\mathrm{means}\:{r}×{r}_{{a}} ? \\ $$ Commented…
Question Number 76090 by john santuy last updated on 23/Dec/19 Commented by Prithwish sen last updated on 23/Dec/19 $$\mathrm{0} \\ $$ Answered by MJS last…
Question Number 10555 by paonky last updated on 17/Feb/17 $$\mathrm{why}\:\Gamma\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \left[−\mathrm{ln}\left(\mathrm{u}\right)\right]^{{x}−\mathrm{1}} {du}\:\:? \\ $$$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{how}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{this} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 141627 by cherokeesay last updated on 21/May/21 Answered by MJS_new last updated on 21/May/21 $$\frac{\mathrm{sin}\:\mathrm{2}{x}\:+\mathrm{sin}\:\mathrm{4}{x}}{\mathrm{1}+\mathrm{cos}\:\mathrm{2}{x}\:+\mathrm{cos}\:\mathrm{4}{x}}=\frac{−\mathrm{2sin}\:{x}\:\mathrm{cos}\:{x}\:\left(\mathrm{1}−\mathrm{4cos}^{\mathrm{2}} \:{x}\right)}{\mathrm{1}+\mathrm{2cos}^{\mathrm{2}} \:{x}\:\left(\mathrm{1}−\mathrm{4sin}^{\mathrm{2}} \:{x}\right)}= \\ $$$$\:\:\:\:\:\left[\mathrm{sin}\:{x}\:=\frac{\mathrm{tan}\:{x}}{\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{x}}}\wedge\mathrm{cos}\:{x}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{x}}}\right] \\…
Question Number 76088 by aliesam last updated on 23/Dec/19 Commented by mr W last updated on 23/Dec/19 $${r}=\mathrm{5} \\ $$$${see}\:{Q}#\mathrm{64112} \\ $$ Commented by aliesam…
Question Number 10553 by shiv ram last updated on 17/Feb/17 $$\left(\mathrm{D}^{\mathrm{2}} +\mathrm{4}\right)\mathrm{y}=\mathrm{tan}\:\mathrm{2x}\:\:\:\:\:\:\:\:\:\:\:\mathrm{D}=\mathrm{d}/\mathrm{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com