Consider a function of two variables $h(x,y)$.
If it's linear for $y$. Can I express it as $a(x)y + b(x)$ ? If positive, why ?
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$\begingroup$ Short answer: Yes. $\endgroup$– vonbrandMar 15, 2013 at 14:09
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$\begingroup$ I added the original sentence "If positive, why ?", although I don't understand what you have in mind. $\endgroup$– Américo TavaresMar 15, 2013 at 14:16
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$\begingroup$ I asked if i indeed could express as a(x)y + b(x) what would be the reason that wouldallow me to. $\endgroup$– nerdyMar 15, 2013 at 14:20
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$\begingroup$ I'm doing electrical engineer too.We learnt in Signals and Systems that for a system to be linear the resopnse to the sum of two different inputs must be the sum of the response of the two different inputs separately.In the case mentioned y = mx + c is not a linear system.Why is "h(x) = a(x)y + b(x)" ( considering m=a(x) and c=b(x) ) linear ? $\endgroup$– nerdyMar 15, 2013 at 14:22
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2 Answers
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Depending on your source, "linear for $y$" may mean different things. It may mean that $h$ is a linear map in the variable $y$, so that $$h(x,y_1+y_2)=h(x,y_1)+h(x,y_2)$$ and $$h(x,\alpha\cdot y)=\alpha\cdot h(x,y)$$ whenever both sides make sense. In that case--and if $h$ is a function $\Bbb R^2\to\Bbb R$, say--we do, indeed, have $$h(x,y)=a(x)\cdot y+b(x),$$ with $a(x)=h(x,1)$ and $b(x)\equiv 0$.
It could also be meaning to denote an affine map, in which case--again assuming $h:\Bbb R^2\to\Bbb R$--we have $$h(x,y)=a(x)\cdot y+b(x),$$ with $a(x)=h(x,1)$ and $b(x)=h(x,0)$.
answered Mar 15, 2013 at 14:27
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$\begingroup$ Cameron, do you recommend me any video lecture, article, book or site to better understand the "maps" you mentioned ? I have clear interest since i'm confused with my Signals and Systems ( electrical engineer ) notation of linear, which i guess is "linear map notation", and the affine map notation for linear. $\endgroup$– nerdyMar 15, 2013 at 14:33
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$\begingroup$ This page has a pretty good intuitive guide to linear maps. As for confusion with affine maps, just remember that $f(x)=mx+b$ has a graph that is a line, but it is only a linear map when $b=0$. It's a pain, but you might want to just forget that you ever thought of such a function as linear. $\endgroup$ Mar 15, 2013 at 14:56
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$\begingroup$ Lol! I knew this amazing page ( amazing approach for complex numbers, euler number and sin ) but didn't know there was such an article there.Thanks for the insight Cameron $\endgroup$– nerdyMar 15, 2013 at 15:03
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A linear function would have the form
$$y=m.x+c$$
If its linear for y then $$h(x,y) = m.y+c\tag1$$
Now as this function is also dependent on $x$ so it would mean, $$m=f(x);C=\phi(x)\tag2$$
Substituting in (2) in (1)
$$h(x,y) = f(x).y+\phi(x)$$
So this is the form that you were intending to represent
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$\begingroup$ Why f(x) = y = mx + c is linear for x ? Let's consider this function as a system, x would be the input and y the output.If the system is linear then the sum of the response of two different inputs must be the response of the sum of this two different inputs.But thats not the case : considerar x1 the first input and x2 the second input: y1 = mx1 + c // y2 = mx2 + c // y3 = m(x1 + x2) + c // In this case y3 is clearly different from y1 + y2. $\endgroup$– nerdyMar 15, 2013 at 14:13