Differential Operators: Understanding the Result of Dx-xD

AI Thread Summary
The discussion centers on the differential operator expression (D-x)(D+x) and the resulting term Dx-xD, which is asserted to equal 1. Participants clarify that the relationship arises from the product rule of differentiation, where Dx can be expressed as 1 + xD. This leads to the conclusion that (Dx - xD)f(x) simplifies to f(x), confirming that Dx - xD equals 1. Additionally, it is noted that -Dx + xD indeed results in -1. Understanding these relationships is essential for grasping differential operators in the context of differential equations.
tirnanog84
Messages
2
Reaction score
0
We're doing differential operators in my Differential Equations class right now, and our professor assigned the following problem to us:

(D-x)(D+x)

Which inevitably gives us the following terms as part of the final answer: Dx-xD

The answer in the book tells me that Dx-xD = 1, and some preliminary research has told me that this is true. What I couldn't find was the why. Why, or how, does Dx-xD result in 1? And does -Dx+xD = -1?
 
Physics news on Phys.org
tirnanog84 said:
We're doing differential operators in my Differential Equations class right now, and our professor assigned the following problem to us:

(D-x)(D+x)

Which inevitably gives us the following terms as part of the final answer: Dx-xD

The answer in the book tells me that Dx-xD = 1, and some preliminary research has told me that this is true. What I couldn't find was the why. Why, or how, does Dx-xD result in 1? And does -Dx+xD = -1?
product rule gives
Dx=1+xD
might be easier to see with a function
D(xy)=(Dx)y+x(Dy)=y+x(Dy)=(1+xD)y
so
Dx=1+xD
 
That is because
\frac{d}{dx}\left(xf(x)\right)-x\frac{d}{dx}\left(f(x)\right)=xf'(x)+f(x)-xf'(x)=f(x)
(Dx-xD)f(x)=f(x)
Hence Dx-xD=1
 
Thanks! o:)
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top