Question Number 204230 by Davidtim last updated on 09/Feb/24
$${what}\:{does}\:{mean}\:{that}\:{we}\:{say}\:{C}^{°} ={F}^{°} \\ $$$${in}\:−\mathrm{40}? \\ $$
Commented by mr W last updated on 10/Feb/24
$${generally}\:{the}\:{temperature}\:{value}\:{in} \\ $$$${unit}\:°{C}\:\left({Celsius}\right)\:{is}\:{not}\:{the}\:{same}\:{as}\: \\ $$$${its}\:{value}\:{in}\:{unit}\:°{F}\:\left({Fahrenheit}\right),\: \\ $$$${for}\:{example}\:\mathrm{10}°{C}\:=\:\mathrm{50}°{F},\: \\ $$$$−\mathrm{20}°{C}=−\mathrm{4}°{F}.\:{but}\:−\mathrm{40}°{C}=−\mathrm{40}°{F}. \\ $$
Commented by Davidtim last updated on 10/Feb/24
$${thanks} \\ $$
Commented by Davidtim last updated on 10/Feb/24
$${would}\:{you}\:{mind}\:{find}\:{the}\:\frac{{d}}{{dx}}\left({x}!\right)? \\ $$
Commented by mr W last updated on 10/Feb/24
$${x}!=\Gamma\left({x}+\mathrm{1}\right) \\ $$$$\frac{{d}\left({x}!\right)}{{dx}}={x}!\psi^{\left(\mathrm{0}\right)} \left({x}+\mathrm{1}\right) \\ $$
Commented by Davidtim last updated on 12/Feb/24
$${would}\:{you}\:{mind}\:{to}\:{solve}\:{without}\:{gamma} \\ $$$${or}\:{bitta}\:{function}? \\ $$
Commented by mr W last updated on 12/Feb/24
$${i}\:{can}\:{not}.\:{if}\:{you}\:{can},\:{please}\:{share}. \\ $$
Commented by Davidtim last updated on 12/Feb/24
![y=x! lny=ln(x!) lny=ln[x∙(x−1)∙(x−2)∙(x−3)∙∙∙3∙2∙1] lny=lnx+ln(x−1)+ln(x−2)+ln(x−3)+∙∙∙+ln1 (y^′ /y)=(1/x)+(1/(x−1))+(1/(x−2))+∙∙∙+0 y^′ =y((1/x)+(1/(x−1))+(1/(x−2))+∙∙) y^′ =x!((1/x)+(1/(x−1))+(1/(x−2))+∙∙∙)](https://www.tinkutara.com/question/Q204325.png)
$${y}={x}! \\ $$$${lny}={ln}\left({x}!\right) \\ $$$${lny}={ln}\left[{x}\centerdot\left({x}−\mathrm{1}\right)\centerdot\left({x}−\mathrm{2}\right)\centerdot\left({x}−\mathrm{3}\right)\centerdot\centerdot\centerdot\mathrm{3}\centerdot\mathrm{2}\centerdot\mathrm{1}\right] \\ $$$${lny}={lnx}+{ln}\left({x}−\mathrm{1}\right)+{ln}\left({x}−\mathrm{2}\right)+{ln}\left({x}−\mathrm{3}\right)+\centerdot\centerdot\centerdot+{ln}\mathrm{1} \\ $$$$\frac{{y}^{'} }{{y}}=\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{x}−\mathrm{1}}+\frac{\mathrm{1}}{{x}−\mathrm{2}}+\centerdot\centerdot\centerdot+\mathrm{0} \\ $$$${y}^{'} ={y}\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{x}−\mathrm{1}}+\frac{\mathrm{1}}{{x}−\mathrm{2}}+\centerdot\centerdot\right) \\ $$$${y}^{'} ={x}!\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{x}−\mathrm{1}}+\frac{\mathrm{1}}{{x}−\mathrm{2}}+\centerdot\centerdot\centerdot\right) \\ $$
Commented by Davidtim last updated on 12/Feb/24
![y=x! lny=ln(x!) lny=ln[x∙(x−1)∙(x−2)∙(x−3)∙∙∙3∙2∙1] lny=lnx+ln(x−1)+ln(x−2)+ln(x−3)+∙∙∙+ln1 (y^′ /y)=(1/x)+(1/(x−1))+(1/(x−2))+∙∙∙+0 y^′ =y((1/x)+(1/(x−1))+(1/(x−2))+∙∙) y^′ =x!((1/x)+(1/(x−1))+(1/(x−2))+∙∙∙)](https://www.tinkutara.com/question/Q204326.png)
$${y}={x}! \\ $$$${lny}={ln}\left({x}!\right) \\ $$$${lny}={ln}\left[{x}\centerdot\left({x}−\mathrm{1}\right)\centerdot\left({x}−\mathrm{2}\right)\centerdot\left({x}−\mathrm{3}\right)\centerdot\centerdot\centerdot\mathrm{3}\centerdot\mathrm{2}\centerdot\mathrm{1}\right] \\ $$$${lny}={lnx}+{ln}\left({x}−\mathrm{1}\right)+{ln}\left({x}−\mathrm{2}\right)+{ln}\left({x}−\mathrm{3}\right)+\centerdot\centerdot\centerdot+{ln}\mathrm{1} \\ $$$$\frac{{y}^{'} }{{y}}=\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{x}−\mathrm{1}}+\frac{\mathrm{1}}{{x}−\mathrm{2}}+\centerdot\centerdot\centerdot+\mathrm{0} \\ $$$${y}^{'} ={y}\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{x}−\mathrm{1}}+\frac{\mathrm{1}}{{x}−\mathrm{2}}+\centerdot\centerdot\right) \\ $$$${y}^{'} ={x}!\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{x}−\mathrm{1}}+\frac{\mathrm{1}}{{x}−\mathrm{2}}+\centerdot\centerdot\centerdot\right) \\ $$
Commented by mr W last updated on 12/Feb/24
$${when}\:{you}\:{take}\:{x}!={x}\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)…\mathrm{3}×\mathrm{2}×\mathrm{1} \\ $$$${you}\:{assume}\:{x}\in{N},\:{then}\:{x}!\:{is}\:{not} \\ $$$${a}\:{continuous}\:{function}\:{and}\:{thus}\:{not} \\ $$$${differentiable}. \\ $$
Commented by Davidtim last updated on 13/Feb/24
$${how}\:{can}\:{solve}\:{without}\:{gamma}\:{function}? \\ $$
Answered by Satyam1234 last updated on 09/Feb/24
$${F}={C}×\frac{\mathrm{9}}{\mathrm{5}}+\mathrm{32} \\ $$$${Let}\:{C}={F}={x} \\ $$$${x}=\frac{\mathrm{9}{x}}{\mathrm{5}}+\mathrm{32} \\ $$$$\mathrm{5}{x}=\mathrm{9}{x}+\mathrm{160} \\ $$$$−\mathrm{4}{x}=\mathrm{160} \\ $$$${x}=−\mathrm{40} \\ $$
Commented by Davidtim last updated on 09/Feb/24
$${I}\:{don}'{t}\:{understant}\:{the}\:{main}\:{mean}\:{ot} \\ $$$${this}. \\ $$$${describe}\:{the}\:{main}\:{mean}\:{of}\:{the}\:−\mathrm{40}. \\ $$
Commented by JDamian last updated on 10/Feb/24
$$\mathrm{the}\:\mathrm{main}\:\mathrm{purpose}\:\mathrm{is}\:\mathrm{to}\:\mathrm{make}\:\mathrm{you}\:\mathrm{to}\:\mathrm{solve} \\ $$$$\mathrm{a}\:\mathrm{first}\:\mathrm{degree}\:\mathrm{equation} \\ $$