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Question Number 33688 by mondodotto@gmail.com last updated on 22/Apr/18
given that f(x)=(1/2)(10^x +10^(−x) ) prove that 2f(x) f(y)=f(x+y)+f(x−y)
$$\boldsymbol{\mathrm{given}}\:\boldsymbol{\mathrm{that}}\: \\ $$$$\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{10}^{\boldsymbol{{x}}} +\mathrm{10}^{−\boldsymbol{{x}}} \right)\:\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{that}} \\ $$$$\mathrm{2}\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{y}}\right)=\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}\right)+\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{y}}\right) \\ $$
Answered by Rasheed.Sindhi last updated on 22/Apr/18
f(x)=(1/2)(10^x +10^(−x) ) f(y)=(1/2)(10^y +10^(−y) ) 2f(x) f(y)=2×(1/2)(10^x +10^(−x) )×(1/2)(10^y +10^(−y) ) =(1/2)(10^(x+y) +10^(−x−y) +10^(x−y) +10^(y−x) ) =(1/2)(10^(x+y) +10^(−(x+y)) +10^(x−y) +10^(−(x−y)) ) =(1/2) (10^(x+y) +10^(−(x+y)) )+(1/2)(10^(x−y) +10^(−(x−y)) ) =f(x+y)+f(x−y)
$$\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{10}^{\boldsymbol{{x}}} +\mathrm{10}^{−\boldsymbol{{x}}} \right) \\ $$$$\boldsymbol{\mathrm{f}}\left(\mathrm{y}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{10}^{\mathrm{y}} +\mathrm{10}^{−\mathrm{y}} \right) \\ $$$$\mathrm{2}\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{y}}\right)=\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{10}^{\boldsymbol{{x}}} +\mathrm{10}^{−\boldsymbol{{x}}} \right)×\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{10}^{\mathrm{y}} +\mathrm{10}^{−\mathrm{y}} \right) \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{10}^{\mathrm{x}+\mathrm{y}} +\mathrm{10}^{−\mathrm{x}−\mathrm{y}} +\mathrm{10}^{\mathrm{x}−\mathrm{y}} +\mathrm{10}^{\mathrm{y}−\mathrm{x}} \right) \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{10}^{\mathrm{x}+\mathrm{y}} +\mathrm{10}^{−\left(\mathrm{x}+\mathrm{y}\right)} +\mathrm{10}^{\mathrm{x}−\mathrm{y}} +\mathrm{10}^{−\left(\mathrm{x}−\mathrm{y}\right)} \right) \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\:\left(\mathrm{10}^{\mathrm{x}+\mathrm{y}} +\mathrm{10}^{−\left(\mathrm{x}+\mathrm{y}\right)} \right)+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{10}^{\mathrm{x}−\mathrm{y}} +\mathrm{10}^{−\left(\mathrm{x}−\mathrm{y}\right)} \right) \\ $$$$\:\:\:\:=\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}\right)+\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{y}}\right) \\ $$

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