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Question Number 188443 Answers: 1 Comments: 0

Question Number 188441 Answers: 1 Comments: 0

Question Number 188442 Answers: 1 Comments: 0

Solve by computer programing (if possible) d<a<b & c < a, b>2c a^2 +b^2 = 5c^2 +2d^2 (a, b, c, d ∈ N) c^2 +d^2 = a^2 .........(i) (2c)^2 +d^2 = b^2 .........(ii) (a, b, c, d) = ?

$$ \\ $$$${Solve}\:{by}\:{computer}\:{programing} \\ $$$$\left({if}\:{possible}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{d}<{a}<{b}\:\&\:{c}\:<\:{a},\:{b}>\mathrm{2}{c} \\ $$$$\cancel{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\:\mathrm{5}{c}^{\mathrm{2}} +\mathrm{2}{d}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\:\:\:\left({a},\:{b},\:{c},\:{d}\:\:\in\:\mathrm{N}\right) \\ $$$${c}^{\mathrm{2}} +{d}^{\mathrm{2}} \:=\:{a}^{\mathrm{2}} \:\:\:\:\:\:\:\:.........\left({i}\right) \\ $$$$\left(\mathrm{2}{c}\right)^{\mathrm{2}} +{d}^{\mathrm{2}} \:=\:{b}^{\mathrm{2}} \:\:.........\left({ii}\right) \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({a},\:{b},\:{c},\:{d}\right)\:=\:? \\ $$

Question Number 188430 Answers: 0 Comments: 0

2⌊ x ⌋ − ⌊ −x ⌋ =4 −−−− if x∈Z ⇒ 2x +x = 4 ⇒ x=(4/3) ,impossible if x∉ Z ⇒^(⌊−x⌋=−⌊x⌋−1) 2⌊x⌋+⌊x⌋=3 ⇒ ⌊ x ⌋= 1 ⇒ 1≤ x < 2 ⇒^(x≠1) x∈ (1 , 2) ✓

$$ \\ $$$$\:\:\:\:\mathrm{2}\lfloor\:{x}\:\rfloor\:−\:\lfloor\:−{x}\:\rfloor\:=\mathrm{4} \\ $$$$\:\:\:−−−− \\ $$$$\:\:{if}\:\:{x}\in\mathbb{Z}\:\Rightarrow\:\:\mathrm{2}{x}\:+{x}\:=\:\mathrm{4}\:\Rightarrow\:{x}=\frac{\mathrm{4}}{\mathrm{3}}\:\:,{impossible} \\ $$$$\:\:{if}\:{x}\notin\:\mathbb{Z}\:\overset{\lfloor−{x}\rfloor=−\lfloor{x}\rfloor−\mathrm{1}} {\Rightarrow}\mathrm{2}\lfloor{x}\rfloor+\lfloor{x}\rfloor=\mathrm{3} \\ $$$$\:\:\:\:\:\Rightarrow\:\lfloor\:{x}\:\rfloor=\:\mathrm{1}\:\Rightarrow\:\:\mathrm{1}\leqslant\:{x}\:<\:\mathrm{2}\:\:\:\:\overset{{x}\neq\mathrm{1}} {\Rightarrow}\:{x}\in\:\left(\mathrm{1}\:,\:\mathrm{2}\right)\:\:\:\checkmark \\ $$$$ \\ $$

Question Number 188418 Answers: 2 Comments: 0

Question Number 188417 Answers: 1 Comments: 1

Question Number 188416 Answers: 1 Comments: 0

Question Number 188407 Answers: 3 Comments: 0

xf(x) = f(x + 2) f(2) = 2 f(8) = ?

$${xf}\left({x}\right)\:=\:{f}\left({x}\:+\:\mathrm{2}\right) \\ $$$${f}\left(\mathrm{2}\right)\:=\:\mathrm{2} \\ $$$${f}\left(\mathrm{8}\right)\:=\:?\: \\ $$

Question Number 188503 Answers: 0 Comments: 0

Question Number 188392 Answers: 2 Comments: 3

Question Number 188387 Answers: 2 Comments: 0

Question Number 188384 Answers: 1 Comments: 0

if ∫_0 ^∞ e^(−ax) dx = (1/a) show that ∫_0 ^∞ e^(−ax) x^n dx = ((n!)/a^(n+1) )

$$\:\:\: \\ $$$$\:\:\:\mathrm{if}\:\:\:\:\int_{\mathrm{0}} ^{\infty} {e}^{−{ax}} {dx}\:\:=\:\:\frac{\mathrm{1}}{{a}} \\ $$$$\:\:\:\:\:\:\:\:{show}\:{that}\:\:\int_{\mathrm{0}} ^{\infty} {e}^{−{ax}} \:{x}^{{n}} {dx}\:\:=\:\:\frac{{n}!}{{a}^{{n}+\mathrm{1}} } \\ $$

Question Number 188381 Answers: 2 Comments: 0

Advanced calculus Find the value of the following series. Ω = Σ_(n=1) ^∞ (( (−1)^( n) ζ ( n ))/(n. 2^( n) )) = ? ζ ( z ) = Σ_( n=1) ^∞ (( 1)/(n^( z) )) ; Re ( z )>1

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{Advanced}\:\:\mathrm{calculus} \\ $$$$ \\ $$$$\:\:\:\:\:\mathrm{Find}\:\:\mathrm{the}\:\:\mathrm{value}\:\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{series}. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\Omega\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\:\left(−\mathrm{1}\right)^{\:{n}} \:\zeta\:\left(\:{n}\:\right)}{{n}.\:\mathrm{2}^{\:{n}} }\:=\:? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\zeta\:\left(\:{z}\:\right)\:=\:\underset{\:{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\mathrm{1}}{{n}^{\:{z}} \:}\:\:\:\:\:;\:\:\:\:\mathscr{R}{e}\:\left(\:{z}\:\right)>\mathrm{1} \\ $$$$ \\ $$

Question Number 188380 Answers: 1 Comments: 0