Mapping Notation

cscott · Sep 26, 2005

Mapping Rule

Say I have the function [itex]y = 2 \sin 3(x - 20)[/itex] and the corresponding mapping notation [itex] (x, y) \rightarrow (\frac{1}{3}x + 20, 2y)[/itex] (which I assume is correct.) How come I take the inverse of the amplitude (2) and horizontal "compression" (3), and how come a negative phase shift moves the wave to the right? What is the true purpose of mapping notation?

edit: I guess this is more properly called "mapping rule," true?

Thanks.

HallsofIvy · Sep 26, 2005

You are saying, I think, that x'= (1/3)x+20 and y'= 2y. Okay, then x= 3(x'-20)= 3x- 60 and y= y'/2. Substituting those, y= 2sin(3(x- 20)) becomes y'/2= 2 sin(x') or y'= 4sin(x') (Did you intend y-> y/2 rather than y-> 2y? That would give y'= sin(x'), much simpler).

The amplitude of y'= 4sin(x') is, of course, 4. I have absolutely no idea what you mean by "inverse of amplitude". The amplitude is a number and perhaps by "inverse" you mean "reciprocal. The point is, as I stated above, that gives Ay'= Asin(...) and so the A's cancel.

"how come a negative phase shift moves the wave to the right". Well, any y= sin(x- b) is 0 when x- b= 0 which is the same as saying x= b. That is, the graph is move from x= 0 to x= b.

cscott · Sep 26, 2005

I've never had it explained like that before but I think I get what you mean.

When I said "inverse" I did mean "reciprocal," sorry.

(Did you intend y-> y/2 rather than y-> 2y? That would give y'= sin(x'), much simpler).

According to my teacher, an amplitude of x (in standard form) will be 1/x in my mapping rule, but I see it doesn't work out right the way you did it. Is something wrong?

Am I right in saying the the mapping rule maps the function you're working with to sin(x)?

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Mapping Notation

cscott

HallsofIvy

cscott