Graphical transformation

Question

I have a burning question to ask regarding graphical transformation:

Suppose I have a function $f(x)$

I want to find $f(ax+b)$ for non zero $a,b$.

There are two approaches that I can go:

First: $f(x)\mapsto f(x+\frac{b}{a})\mapsto f(ax+b)$

Second: $f(x)\mapsto f(ax)\mapsto f(ax+b)$

However, when I start to plot the graphical transformation from $\sin(x)$ to $\sin(2x+\pi)$, the first approach will get me a wrong answer. (I will obtain $\cos(2x)$ instead of $\sin(2x)$)

Please help. Thank you.

Answer 1

The mistake you made, basically, was knowing too much and being too eager.

When applying the first approach, you did:

$$\sin(x)\mapsto \sin(x+\frac{\pi}{2})=\cos(x)\mapsto\cos(2x)$$

while you should do this:

$$\sin(x) \mapsto \sin(x+\frac\pi2)\mapsto \sin(2x+\pi)$$

Your mistake was thinking that just because, for a pair $x,y$ (in your particular case, $y$ was equal to $x+\pi/2$) you have $\sin(y) = \cos(x)$, you cannot assume that you also have $\sin(2y) = \cos(2x)$, which is where you made your mistake. You assumed (implicitly) that that is the case, by saying

It's no differetnt to multiply the argument of $\sin(y)$ by $2$ as it is to multiply the argument of $\cos(x)$ by $2$.

Answer 2

After thinking about the responses above and a chat with some of my seniors, here is something that I will like to make:

$f(x)\mapsto f(ax):=g(x)\mapsto g(x+\frac{b}{a})=f(ax+b)---(1)$ $f(x)\mapsto f(x+b):=h(x)\mapsto h(ax)=f(ax+b)---(2)$

For the case of $\sin(x)\mapsto \sin(2x+\pi)$,

Using method $(1)$: $\sin(x)\mapsto \sin(2x):=g(x)\mapsto g(x+\frac{\pi}{2})=\sin(2x+\pi)$

Using method $(2)$: $\sin(x)\mapsto \sin(x+\pi):=h(x)\mapsto h(2x)=\sin(2x+\pi)$

Method $(1)$ involves a scaling along $x$ axis, followed by a phase change of $\frac{\pi}{2}$.

Method $(1)$ involves a phase change of $\pi$, followed by a scaling along $x$ axis.

The question I posed earlier, especially at $$f(x)\mapsto f(x+\frac{b}{a})\mapsto f(ax+b)$$ treats $x+\frac{b}{a}$ as a whole entity.

I believe if we change the $x$ axis to $x+\frac{b}{a}$ axis, then we can do the have the scaling of factor $a$ to make it $$f[a(x+\frac{b}{a})]=f(ax+b)$$ I don't just simply replace $x$ by $ax$, which otherwise I will get $f(ax+\frac{b}{a})$.