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Question Number 228076 Answers: 3 Comments: 2

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x+13=22

$${x}+\mathrm{13}=\mathrm{22} \\ $$

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I = ∫ (1/(x^5 +1)) dx = ∫ (1/((x+1)(x^4 −x^3 +x^2 −x+1))) dx = ∫ [(1/(5(x+1))) − ((x^3 −2x^2 +3x−4)/(5(x^4 −x^3 +x^2 −x+1)))] dx = (1/5)∫ (1/(x+1)) dx − (1/5)∫ ((x^3 −2x^2 +3x−4)/(x^4 −x^3 +x^2 −x+1)) dx = (1/5)ln ∣x+1∣ + C_1 − (1/5)∫ ((x^3 −2x^2 +3x−4)/(x^4 −x^3 +x^2 −x+1)) dx I_2 = ∫ ((x^3 −2x^2 +3x−4)/(x^4 −x^3 +x^2 −x+1)) dx = ∫ ((x^3 −2x^2 +3x−4)/((x^2 +(ϕ−1)x+1)(x^2 −ϕx+1))) dx = ∫ [(((1−ϕ)x−2)/(x^2 +(ϕ−1)x+1)) + ((ϕx−2)/(x^2 −ϕx+1))] dx = ∫ [(((1−ϕ)x−2)/(x^2 +(ϕ−1)x+1))] dx + ∫ [((ϕx−2)/(x^2 −ϕx+1))] dx = ((1−ϕ)/2)ln ∣x^2 +(ϕ−1)x+1∣ − (√(2−ϕ))tan^(−1) (((2x+ϕ−1)/( (√(2+ϕ))))) − (ϕ/2)ln ∣x^2 −ϕx+1∣ − (√(3−ϕ))tan^(−1) (((2x−ϕ)/( (√(3−ϕ))))) + C_2 I = determinant ((((1/5)ln ∣x+1∣ − ((1−ϕ)/(10))ln ∣x^2 +(ϕ−1)x+1∣ + (√(2−ϕ))tan^(−1) (((2x+ϕ−1)/( (√(2+ϕ))))) + (ϕ/(10))ln ∣x^2 −ϕx+1∣ + ((√(3−ϕ))/5)tan^(−1) (((2x−ϕ)/( (√(3−ϕ))))) + C))) Where ϕ = Golden Ratio = (((√5) + 1)/2) I wonder if I made any mistakes here. Can someone check this?

$${I}\:=\:\int\:\frac{\mathrm{1}}{{x}^{\mathrm{5}} +\mathrm{1}}\:{dx}\:=\:\int\:\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{4}} −{x}^{\mathrm{3}} +{x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}\:{dx} \\ $$$$=\:\int\:\left[\frac{\mathrm{1}}{\mathrm{5}\left({x}+\mathrm{1}\right)}\:−\:\frac{{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{4}}{\mathrm{5}\left({x}^{\mathrm{4}} −{x}^{\mathrm{3}} +{x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}\right]\:{dx} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{5}}\int\:\frac{\mathrm{1}}{{x}+\mathrm{1}}\:{dx}\:−\:\frac{\mathrm{1}}{\mathrm{5}}\int\:\frac{{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{4}}{{x}^{\mathrm{4}} −{x}^{\mathrm{3}} +{x}^{\mathrm{2}} −{x}+\mathrm{1}}\:{dx} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{5}}\mathrm{ln}\:\mid{x}+\mathrm{1}\mid\:+\:{C}_{\mathrm{1}} \:−\:\frac{\mathrm{1}}{\mathrm{5}}\int\:\frac{{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{4}}{{x}^{\mathrm{4}} −{x}^{\mathrm{3}} +{x}^{\mathrm{2}} −{x}+\mathrm{1}}\:{dx} \\ $$$$ \\ $$$${I}_{\mathrm{2}} \:=\:\int\:\frac{{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{4}}{{x}^{\mathrm{4}} −{x}^{\mathrm{3}} +{x}^{\mathrm{2}} −{x}+\mathrm{1}}\:{dx}\:=\:\int\:\frac{{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{4}}{\left({x}^{\mathrm{2}} +\left(\varphi−\mathrm{1}\right){x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −\varphi{x}+\mathrm{1}\right)}\:{dx} \\ $$$$=\:\int\:\left[\frac{\left(\mathrm{1}−\varphi\right){x}−\mathrm{2}}{{x}^{\mathrm{2}} +\left(\varphi−\mathrm{1}\right){x}+\mathrm{1}}\:+\:\frac{\varphi{x}−\mathrm{2}}{{x}^{\mathrm{2}} −\varphi{x}+\mathrm{1}}\right]\:{dx} \\ $$$$=\:\int\:\left[\frac{\left(\mathrm{1}−\varphi\right){x}−\mathrm{2}}{{x}^{\mathrm{2}} +\left(\varphi−\mathrm{1}\right){x}+\mathrm{1}}\right]\:{dx}\:+\:\int\:\left[\frac{\varphi{x}−\mathrm{2}}{{x}^{\mathrm{2}} −\varphi{x}+\mathrm{1}}\right]\:{dx} \\ $$$$=\:\frac{\mathrm{1}−\varphi}{\mathrm{2}}\mathrm{ln}\:\mid{x}^{\mathrm{2}} +\left(\varphi−\mathrm{1}\right){x}+\mathrm{1}\mid\:−\:\sqrt{\mathrm{2}−\varphi}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}+\varphi−\mathrm{1}}{\:\sqrt{\mathrm{2}+\varphi}}\right)\:−\:\frac{\varphi}{\mathrm{2}}\mathrm{ln}\:\mid{x}^{\mathrm{2}} −\varphi{x}+\mathrm{1}\mid\:−\:\sqrt{\mathrm{3}−\varphi}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}−\varphi}{\:\sqrt{\mathrm{3}−\varphi}}\right)\:+\:{C}_{\mathrm{2}} \\ $$$$ \\ $$$${I}\:=\:\begin{array}{|c|}{\frac{\mathrm{1}}{\mathrm{5}}\mathrm{ln}\:\mid{x}+\mathrm{1}\mid\:−\:\frac{\mathrm{1}−\varphi}{\mathrm{10}}\mathrm{ln}\:\mid{x}^{\mathrm{2}} +\left(\varphi−\mathrm{1}\right){x}+\mathrm{1}\mid\:+\:\sqrt{\mathrm{2}−\varphi}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}+\varphi−\mathrm{1}}{\:\sqrt{\mathrm{2}+\varphi}}\right)\:+\:\frac{\varphi}{\mathrm{10}}\mathrm{ln}\:\mid{x}^{\mathrm{2}} −\varphi{x}+\mathrm{1}\mid\:+\:\frac{\sqrt{\mathrm{3}−\varphi}}{\mathrm{5}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}−\varphi}{\:\sqrt{\mathrm{3}−\varphi}}\right)\:+\:{C}}\\\hline\end{array} \\ $$$$\mathrm{Where}\:\varphi\:=\:\mathrm{Golden}\:\mathrm{Ratio}\:=\:\frac{\sqrt{\mathrm{5}}\:+\:\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{wonder}\:\mathrm{if}\:\mathrm{I}\:\mathrm{made}\:\mathrm{any}\:\mathrm{mistakes}\:\mathrm{here}. \\ $$$$\mathrm{Can}\:\mathrm{someone}\:\mathrm{check}\:\mathrm{this}? \\ $$

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t is the fractional part of a, and a^2 +t^2 =18. find t=?

$${t}\:{is}\:{the}\:{fractional}\:{part}\:{of}\:{a},\:{and} \\ $$$${a}^{\mathrm{2}} +{t}^{\mathrm{2}} =\mathrm{18}.\:{find}\:{t}=? \\ $$

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Question Number 227983 Answers: 2 Comments: 0

(√y)+x=a (√x)+y=a ∀x,y,a∈Z

$$ \\ $$$$\sqrt{\mathrm{y}}+\mathrm{x}=\mathrm{a} \\ $$$$\sqrt{\mathrm{x}}+\mathrm{y}=\mathrm{a} \\ $$$$\forall\mathrm{x},\mathrm{y},\mathrm{a}\in\mathrm{Z} \\ $$$$ \\ $$

Question Number 227982 Answers: 3 Comments: 1

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Q. lim lim_(x→a y→b) A_(m,n) =L Suppose a double sequence A_(m,n) converges to L. According to the Moore-Osgood Theorem the order of limits can be interchanged if at least one direction converges uniformly. I find this counterintuitve I don′t understand why uniform convergence in just one direction is sufficient to guarantee the validity of switching the limits. It would be much easier to accept if uniform convergence were required in both directions.

$$\mathrm{Q}. \\ $$$$\underset{{x}\rightarrow{a}\:{y}\rightarrow{b}} {\mathrm{lim}\:\mathrm{lim}}\:{A}_{{m},{n}} ={L} \\ $$$$\: \\ $$$$\mathrm{Suppose}\:\mathrm{a}\:\mathrm{double}\:\mathrm{sequence}\:{A}_{{m},{n}} \:\mathrm{converges}\:\mathrm{to}\:\mathrm{L}. \\ $$$$\mathrm{According}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{Moore}-\mathrm{Osgood}\:\mathrm{Theorem}\:\mathrm{the}\:\mathrm{order}\:\mathrm{of}\:\mathrm{limits}\:\mathrm{can}\:\mathrm{be}\:\mathrm{interchanged} \\ $$$$\mathrm{if}\:\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:\mathrm{direction}\:\mathrm{converges}\:\mathrm{uniformly}. \\ $$$$\mathrm{I}\:\mathrm{find}\:\mathrm{this}\:\mathrm{counterintuitve} \\ $$$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{understand}\:\mathrm{why}\:\mathrm{uniform}\:\mathrm{convergence}\:\mathrm{in} \\ $$$$\mathrm{just}\:\mathrm{one}\:\mathrm{direction}\:\mathrm{is}\:\mathrm{sufficient}\:\mathrm{to}\:\mathrm{guarantee}\:\mathrm{the}\:\mathrm{validity}\:\mathrm{of}\: \\ $$$$\mathrm{switching}\:\mathrm{the}\:\mathrm{limits}. \\ $$$$\mathrm{It}\:\mathrm{would}\:\mathrm{be}\:\mathrm{much}\:\mathrm{easier}\:\mathrm{to}\:\mathrm{accept}\:\mathrm{if}\:\mathrm{uniform} \\ $$$$\mathrm{convergence}\:\mathrm{were}\:\mathrm{required}\:\mathrm{in}\:\mathrm{both}\:\mathrm{directions}. \\ $$

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