Question-59193

Question Number 59193 by salahahmed last updated on 05/May/19

Answered by MJS last updated on 06/May/19

for x∈N ∫_0 ^∞ t^x e^(−t) dt=x! ⇒ we have one obvious solution at x=5 x^3 −x=x! (x−1)x(x+1)=x! ⇒ x≠0∧x≠1 (x−1)x(x+1)=x(x−1)(x−2)(x−3)... (x+1)=(x−2) ⇒ no solution remember we′re seeking x∈N ⇒ we only have to try factors of the constant (x+1)=(x−2)(x−3) x^2 ...+5=0 ⇒ x=5 (x+1)=(x−2)(x−3)(x−4) x^3 ...−25=0 ⇒ x=5 above x=5 x!>x^3 −x there are solutions for x∈R\Z approximating with a calculator I found x≈1.37439465 (±5×10^(−8) ) there are infinite solutions for x<−3 I found x.≈−3.02 and x≈−3.997 there should be solutions at x=−(2n−1)−δ_n and at x=−2n+ε with n>1∧n∈N and δ_n , ε_n being very small positive numbers

$$\mathrm{for}\:{x}\in\mathbb{N}\:\underset{\mathrm{0}} {\overset{\infty} {\int}}{t}^{{x}} \mathrm{e}^{−{t}} {dt}={x}! \\ $$$$\Rightarrow\:\mathrm{we}\:\mathrm{have}\:\mathrm{one}\:\mathrm{obvious}\:\mathrm{solution}\:\mathrm{at}\:{x}=\mathrm{5} \\ $$$${x}^{\mathrm{3}} −{x}={x}! \\ $$$$\left({x}−\mathrm{1}\right){x}\left({x}+\mathrm{1}\right)={x}! \\ $$$$\Rightarrow\:{x}\neq\mathrm{0}\wedge{x}\neq\mathrm{1} \\ $$$$\left({x}−\mathrm{1}\right){x}\left({x}+\mathrm{1}\right)={x}\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right)… \\ $$$$ \\ $$$$\left({x}+\mathrm{1}\right)=\left({x}−\mathrm{2}\right)\:\Rightarrow\:\mathrm{no}\:\mathrm{solution} \\ $$$$ \\ $$$$\mathrm{remember}\:\mathrm{we}'\mathrm{re}\:\mathrm{seeking}\:{x}\in\mathbb{N}\:\Rightarrow\:\mathrm{we}\:\mathrm{only}\:\mathrm{have} \\ $$$$\mathrm{to}\:\mathrm{try}\:\mathrm{factors}\:\mathrm{of}\:\mathrm{the}\:\mathrm{constant} \\ $$$$\left({x}+\mathrm{1}\right)=\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right) \\ $$$$\:\:\:\:\:{x}^{\mathrm{2}} …+\mathrm{5}=\mathrm{0}\:\Rightarrow\:{x}=\mathrm{5} \\ $$$$\left({x}+\mathrm{1}\right)=\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right)\left({x}−\mathrm{4}\right) \\ $$$$\:\:\:\:\:{x}^{\mathrm{3}} …−\mathrm{25}=\mathrm{0}\:\Rightarrow\:{x}=\mathrm{5} \\ $$$$ \\ $$$$\mathrm{above}\:{x}=\mathrm{5}\:{x}!>{x}^{\mathrm{3}} −{x} \\ $$$$ \\ $$$$\mathrm{there}\:\mathrm{are}\:\mathrm{solutions}\:\mathrm{for}\:{x}\in\mathbb{R}\backslash\mathbb{Z} \\ $$$$\mathrm{approximating}\:\mathrm{with}\:\mathrm{a}\:\mathrm{calculator}\:\mathrm{I}\:\mathrm{found} \\ $$$${x}\approx\mathrm{1}.\mathrm{37439465}\:\left(\pm\mathrm{5}×\mathrm{10}^{−\mathrm{8}} \right) \\ $$$$\mathrm{there}\:\mathrm{are}\:\mathrm{infinite}\:\mathrm{solutions}\:\mathrm{for}\:{x}<−\mathrm{3} \\ $$$$\mathrm{I}\:\mathrm{found}\:{x}.\approx−\mathrm{3}.\mathrm{02}\:\mathrm{and}\:{x}\approx−\mathrm{3}.\mathrm{997} \\ $$$$\mathrm{there}\:\mathrm{should}\:\mathrm{be}\:\mathrm{solutions}\:\mathrm{at}\:{x}=−\left(\mathrm{2}{n}−\mathrm{1}\right)−\delta_{{n}} \\ $$$$\mathrm{and}\:\mathrm{at}\:{x}=−\mathrm{2}{n}+\epsilon\:\mathrm{with}\:{n}>\mathrm{1}\wedge{n}\in\mathbb{N}\:\mathrm{and} \\ $$$$\delta_{{n}} ,\:\epsilon_{{n}} \:\mathrm{being}\:\mathrm{very}\:\mathrm{small}\:\mathrm{positive}\:\mathrm{numbers} \\ $$

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