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Mean Value Theorem

  1. Mar 10, 2009 #1
    1. The problem statement:

    greek letters used here: xi, eta

    The Mean Value theorem applied to f(x,y) = sin(x^2 + y^2) implies with a = 0 and b = 0.

    sin(x^2 + y^2) = 2 xi cos( xi^2 + eta^2)x + 2 eta cos(xi^2 +eta^2) y

    find xi and eta or an accurate approximation to them as a function of x and y.


    2. What i have tried doing:

    Let f(x,y) and its first order partial derivatives be continuos in an open region R and let (a,b) and (x,y) be points in R such that the straight line joining these points lies entirely within R. Then there exists a point (xi, eta) on that line between the endpoints. So we get

    f(x,y) = f (a,b) + f_x (xi, eta) (x-a) + f_y (xi,eta) (y-b)


    ** I am not sure where to go ahead from here, I would just like an idea as to what i can do to find the answer i dont want anyone to solve it as i would love to solve it but i just want someone to shed some light for me to go in the right direction. THANX for all the help.
     
  2. jcsd
  3. Mar 10, 2009 #2
    well, i'm not sure if i'm on the right track with this. i would think of the mean value theorem as a kind of taylor polynomial expansion, where eta and xi represent a number that takes care of the rest of the infinitely many remainder terms. so from here, i would look at the region with center a=0, b=0, and expand about that. since this region contains (0,0), it should contain (0,y) and (x,0) for sufficiently small x and y, arbitrary in this small region.
     
  4. Mar 18, 2009 #3
    Hello, Thanx for the reply. I have been trying to solve that problem using taylor polynomials but i do manage to get half way but i still cannot get the right answer. Is there any other way that anyone can suggest. Or do i need to do anything after doing the taylor polynomials.

    thanks once again
     
  5. Mar 18, 2009 #4
    okay, i was a bit off. i think x and y are fixed and xi and eta are then determined. looking at it, i see no way to solve explicitly. i was thinking maybe choose x and y very small, so that xi and eta are very small so that the cos(xi^2 + eta^2) must be very close to one. you do indicate 'approximation'. also, by symmetry, xi=eta.
     
  6. Mar 19, 2009 #5
    sweet.. thanx for that.. i will try that out now and let u know how it goes.
     
  7. Mar 19, 2009 #6
    no that isn't working. and eta doesnt necessarily equal xi.

    wikipedia says that the only way to do get the mean value theorem to work in two dimensions is to parameterize into one variable and then apply the MVT.

    [edit] I get something like: f(x,y)=grad(f(cx,cy)).(x,y)

    maybe looking at A.B=|A||B|cos(theta) can give an analytic estimate on c.

    [edit2] well...with r^2=x^2+y^2, sin(x)~x+x^3/6, cos(x)~1-x^2/2, and dropping high order terms, i get: c=1/(2r)??
     
    Last edited: Mar 19, 2009
  8. Mar 19, 2009 #7
    i think this works, but i need to be more careful. see if you can confirm the following:

    given: xi=cx, eta=cy for some c in (0,1)

    g(t)=f(tx,ty) = sin((tx)^2+(ty)^2)

    so by mvt in one variable, g(1)-g(0)=g'(c)(1-0) (no such thing as mvt in two var!)

    this equates to: f(x,y)-f(0,0)=grad(f(cx,cy)).(x,y)=grad(f(xi,eta)).(x,y) which is what you have.

    let r^2=x^2+y^2

    we get:

    sin(r^2)=2cx cos((c*r)^2 * x + 2cy cos((c*r)^2 * y
    = 2c r^2 cos( (c*r) ^2 )

    approximating sin and cos up to third order, what else?

    r^2 - r^6/6 = 2c r^2 (1- ((c*r)^2)^2/2) = 2c r^2

    dropping c*r^2*((c*r)^2)^2 which appears to be high order, if c~r is on the same order roughly.

    we can solve this for c, to get: c=1/2 (1 - r^4/6)

    phew, looks like something! maybe simply c=1/2 ?
     
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