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If a.f((x) + b.g(x) = f(x).h(x) + g(x).p(x) As LHS = RHS Then we can conclude that h(x) = a & p(x) = b

Is that right? Can anyone give some explanations? I know that there are more than one solutions can exist, but I'm asking that there is any theorem which conclude this result too?

Engineering Minds
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1 Answers1

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I assume that you are asking the following: "If we have $$a\cdot f(x) + b\cdot g(x) = f(x)\cdot h(x) + g(x)\cdot p(x)$$ can we conclude that $a = h(x) $ and $ b = p(x)$, where $f$, $g$, $h$, $p$ are functions of $ x $ and $ a $ and $ b $ are constants?"

The short answer is no, we cannot assume that, because we don't know that multiplying $f(x) $ by $ h(x) $ won't give us any copies of $ g(x) $, and we don't know that multiplying $ g(x) $ by $ p(x) $ won't give us any copies of $ f(x) $.

For example, what if we had $ f(x) = x$, $g(x) = x^2 $, $ h(x = 1+x $, and $ p(x) = 1 $? Then we have: $$ a\cdot f(x) + b\cdot g(x) = x(x+1) + x^2(1)$$ $$a\cdot x + b\cdot x^2 = x^2 + x + x^2 $$ $$ a\cdot x + b\cdot x^2 = x + 2x^2 $$ Therefore, $ a = 1 $ and $ b = 2 $, which are not equal to what we defined for $ h(x) $ and $ g(x) $.

Jude Quintero
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  • I agree with you but in particular case if the conditions are in favour of doing so.If the eqn is √2(sinx + cosx) = tanx+ cotx => here can we write by Aptitude that √2 = 1/cosx – Engineering Minds Jul 24 '19 at 18:51
  • In that particular case, you can assume that, as long as you are only looking for real solutions (there are imaginary solutions to the equation that obviously wouldn't allow you to assume that). However, what happens in that equation is by no means general – Jude Quintero Jul 24 '19 at 19:14