Calculus question, Diff. Forms?

[DukeSteve]

Homework Statement

Given the radial function F: R_n without zero vector -----> R_n

F(x) = phi(||x||)*x ---- the function value is depends on it's distance from the origin.

Phi is the function : phi: (0,infinity) ---> Real Numbers

How to prove that the diff. form defined by this function is exact?
Please give me a step by step answer!

Homework Equations

The Attempt at a Solution

Please give me a step by step answer, because I don't know where to start. I need a step by step answer please

 

[mighty2000]

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To prove that the differential form defined by the function F is exact, we must show that it is the exterior derivative of another form. In other words, we must find a form G such that dG = F.

Step 1: Understand the problem
First, let's understand what the given function F is doing. It takes a vector x, calculates its distance from the origin ||x||, and then multiplies it by the function phi. This means that F is essentially scaling the vector x by the value of phi at the distance ||x||. Keep this in mind as we move forward.

Step 2: Find the form G
To find the form G, we need to express F as a linear combination of the basis forms in R_n. In this case, the basis forms are dx_1, dx_2, ..., dx_n. Since F is a function from R_n to R_n, it must be expressible as a linear combination of these forms. Let's call the coefficients of this linear combination g_1, g_2, ..., g_n. Then we can write:

F(x) = g_1(x)dx_1 + g_2(x)dx_2 + ... + g_n(x)dx_n

Step 3: Calculate the exterior derivative of G
Now, we can calculate the exterior derivative of G, denoted by dG. Since dG is a linear operator, we can calculate it for each basis form separately. We have:

dG = dg_1(x) ∧ dx_1 + dg_2(x) ∧ dx_2 + ... + dg_n(x) ∧ dx_n

Step 4: Compare dG to F
Now, let's compare this to the original function F. Notice that the coefficient of each basis form in dG is the exterior derivative of the corresponding coefficient in F. In other words, we have:

dG = d(g_1(x)) ∧ dx_1 + d(g_2(x)) ∧ dx_2 + ... + d(g_n(x)) ∧ dx_n

Since phi is a function from (0,infinity) to Real Numbers, its derivative must be defined for all values in this range. This means that we can rewrite the above equation as:

dG = d(phi(||x||)) ∧ x

Step 5: Conclusion
We have now shown that the exterior derivative of G is equal to F