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Let-I-n-0-n-a-coshx-a-sinhx-dx-with-0-lt-a-lt-1-and-define-I-lim-n-I-n-Does-I-exist-




Question Number 4156 by Yozzii last updated on 30/Dec/15
Let I(n)=∫_0 ^n {(√(a+coshx))−(√(a+sinhx))}dx  with 0<a<1 and define I=lim_(n→∞) I(n).  Does I exist?
$${Let}\:{I}\left({n}\right)=\int_{\mathrm{0}} ^{{n}} \left\{\sqrt{{a}+{coshx}}−\sqrt{{a}+{sinhx}}\right\}{dx} \\ $$$${with}\:\mathrm{0}<{a}<\mathrm{1}\:{and}\:{define}\:{I}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}{I}\left({n}\right). \\ $$$${Does}\:{I}\:{exist}?\: \\ $$
Commented by 123456 last updated on 30/Dec/15
lim_(x→∞)  coshx−sinhx=?  lim_(x→∞) (√(a+cosh x))−(√(a+sinh x))=?
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{cosh}{x}−\mathrm{sinh}{x}=? \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\sqrt{{a}+\mathrm{cosh}\:{x}}−\sqrt{{a}+\mathrm{sinh}\:{x}}=? \\ $$
Commented by Yozzii last updated on 01/Jan/16
Suppose (√(a+coshx))−(√(a+sinhx))=0  (√((a+coshx)/(a+sinhx)))=1  coshx=sinhx  ((e^x +e^(−x) )/2)=((e^x −e^(−x) )/2)  e^(−x) =−e^(−x)   2e^(−x) =0 (false)  no real root exists.
$${Suppose}\:\sqrt{{a}+{coshx}}−\sqrt{{a}+{sinhx}}=\mathrm{0} \\ $$$$\sqrt{\frac{{a}+{coshx}}{{a}+{sinhx}}}=\mathrm{1} \\ $$$${coshx}={sinhx} \\ $$$$\frac{{e}^{{x}} +{e}^{−{x}} }{\mathrm{2}}=\frac{{e}^{{x}} −{e}^{−{x}} }{\mathrm{2}} \\ $$$${e}^{−{x}} =−{e}^{−{x}} \\ $$$$\mathrm{2}{e}^{−{x}} =\mathrm{0}\:\left({false}\right) \\ $$$${no}\:{real}\:{root}\:{exists}. \\ $$$$ \\ $$
Commented by Filup last updated on 31/Dec/15
cosh x=((e^x +e^(−x) )/2)  sinh x=((e^x −e^(−x) )/2)    cosh(x)−sinh(x)=((e^x +e^(−x) −e^x +e^(−x) )/2)  =e^(−x)   lim_(x→∞)  cosh(x)−sinh(x) = (1/e^∞ )  =0
$$\mathrm{cosh}\:{x}=\frac{{e}^{{x}} +{e}^{−{x}} }{\mathrm{2}} \\ $$$$\mathrm{sinh}\:{x}=\frac{{e}^{{x}} −{e}^{−{x}} }{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{cosh}\left({x}\right)−\mathrm{sinh}\left({x}\right)=\frac{{e}^{{x}} +{e}^{−{x}} −{e}^{{x}} +{e}^{−{x}} }{\mathrm{2}} \\ $$$$={e}^{−{x}} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{cosh}\left({x}\right)−\mathrm{sinh}\left({x}\right)\:=\:\frac{\mathrm{1}}{{e}^{\infty} } \\ $$$$=\mathrm{0} \\ $$$$ \\ $$
Commented by Filup last updated on 31/Dec/15
Commented by Filup last updated on 31/Dec/15
It seems that  lim_(x→∞)  (√(a+cosh(x)))−(√(a+sinh(x)))=0
$$\mathrm{It}\:\mathrm{seems}\:\mathrm{that} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{{a}+\mathrm{cosh}\left({x}\right)}−\sqrt{{a}+\mathrm{sinh}\left({x}\right)}=\mathrm{0} \\ $$
Commented by Yozzii last updated on 31/Dec/15
I noticed that. So, out of curiosity  I wondered if the area bouned between  the axes for x>0,y>0 and the  curve is finite as n→∞.
$${I}\:{noticed}\:{that}.\:{So},\:{out}\:{of}\:{curiosity} \\ $$$${I}\:{wondered}\:{if}\:{the}\:{area}\:{bouned}\:{between} \\ $$$${the}\:{axes}\:{for}\:{x}>\mathrm{0},{y}>\mathrm{0}\:{and}\:{the} \\ $$$${curve}\:{is}\:{finite}\:{as}\:{n}\rightarrow\infty. \\ $$
Commented by Yozzii last updated on 31/Dec/15
Let             u=a+coshx             v=a+sinhx  g=(√u)−(√v)=((u−v)/( (√u)+(√v)))=((coshx−sinhx)/( (√u)+(√v)))  g=((0.5(e^x +e^(−x) −(e^x −e^(−x) )))/( (√u)+(√v)))=(e^(−x) /( (√u)+(√v)))  g=(1/(e^x ((√(a+0.5(e^x +e^(−x) )))+(√(a+0.5(e^x −e^(−x) ))))))  ∴ lim_(x→∞) g=(1/(∞((√(a+0.5(∞+0)))+(√(a+0.5(∞−0))))))=0.
$${Let} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{u}={a}+{coshx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{v}={a}+{sinhx} \\ $$$${g}=\sqrt{{u}}−\sqrt{{v}}=\frac{{u}−{v}}{\:\sqrt{{u}}+\sqrt{{v}}}=\frac{{coshx}−{sinhx}}{\:\sqrt{{u}}+\sqrt{{v}}} \\ $$$${g}=\frac{\mathrm{0}.\mathrm{5}\left({e}^{{x}} +{e}^{−{x}} −\left({e}^{{x}} −{e}^{−{x}} \right)\right)}{\:\sqrt{{u}}+\sqrt{{v}}}=\frac{{e}^{−{x}} }{\:\sqrt{{u}}+\sqrt{{v}}} \\ $$$${g}=\frac{\mathrm{1}}{{e}^{{x}} \left(\sqrt{{a}+\mathrm{0}.\mathrm{5}\left({e}^{{x}} +{e}^{−{x}} \right)}+\sqrt{{a}+\mathrm{0}.\mathrm{5}\left({e}^{{x}} −{e}^{−{x}} \right)}\right)} \\ $$$$\therefore\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}{g}=\frac{\mathrm{1}}{\infty\left(\sqrt{{a}+\mathrm{0}.\mathrm{5}\left(\infty+\mathrm{0}\right)}+\sqrt{{a}+\mathrm{0}.\mathrm{5}\left(\infty−\mathrm{0}\right)}\right)}=\mathrm{0}. \\ $$

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