how does x operator become x? 6.1 is the equation i wrote down
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how does x operator become x? 6.1 is the equation i wrote down
![equivalence of the two ways of carrying out the transformation.
Example 6.1
Find the operator' obtained by applying a translation through a distance a to the operator. That is
what is the action of ', as defined by Equation 6.6, on an arbitrary f(x)?
Solution: Using the definition of ' (Equation 6.6) and a test function f(x) we have
' f(x) = f(a) x Î (a) f(x),
and since ↑ (a) = (–a) (Equation 6.4),
✰' f(x) = Î(-a) îî (a) f(x).
From Equation 6.12.
Example 6.1. Figure 6.5 illustra
' f(x)=(x + a) f(x).
Finally we may read off the operator
* = x+a.
7(a) y (x) = 4(x-a) = f(x-9
(4₁x)=f(x))
x' f(x) = f(-a) x f(x − a),
and from Equation 6.1 again, Î(-a) [x ƒ(x − a)] = (x + a) f(x), so
299
(6.7)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb1a6ad76-e2c4-4fd8-a398-9f8475c442d4%2Fc21a63fb-35c3-4061-b3dd-05d43bb69a43%2Fw73p21k_processed.png&w=3840&q=75)
Transcribed Image Text:equivalence of the two ways of carrying out the transformation.
Example 6.1
Find the operator' obtained by applying a translation through a distance a to the operator. That is
what is the action of ', as defined by Equation 6.6, on an arbitrary f(x)?
Solution: Using the definition of ' (Equation 6.6) and a test function f(x) we have
' f(x) = f(a) x Î (a) f(x),
and since ↑ (a) = (–a) (Equation 6.4),
✰' f(x) = Î(-a) îî (a) f(x).
From Equation 6.12.
Example 6.1. Figure 6.5 illustra
' f(x)=(x + a) f(x).
Finally we may read off the operator
* = x+a.
7(a) y (x) = 4(x-a) = f(x-9
(4₁x)=f(x))
x' f(x) = f(-a) x f(x − a),
and from Equation 6.1 again, Î(-a) [x ƒ(x − a)] = (x + a) f(x), so
299
(6.7)
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