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Math Help - evaluate

  1. #1
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    evaluate

    Hello, help please

    1) let \tan\alpha and \tan\beta roots of equation x^2 + \pi x + \sqrt{2} = 0 .

    Evaluate : A = \sin^2(\alpha+\beta) + \pi\sin(\alpha+\beta)\cos(\alpha+\beta) + \sqrt{2} \cos^2(\alpha+\beta)

    2) let x,y,z > 0 , prove that : \frac{xy}{z} + \frac{yz}{x} + \frac{zx}{y} \geq x + y + z

    3) Find all function f: \mathbb{R} \to \mathbb{R} such that : f\left( xf(x) + f(y) \right ) = \left(f(x)\right)^2 +y
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  2. #2
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    Quote Originally Posted by linda2005 View Post
    Hello, help please

    1) let \tan\alpha and \tan\beta roots of equation x^2 + \pi x + \sqrt{2} = 0
    [snip]

    You can use the quadratic formula to find a and b (alpha and beta).
    x=(-B+-sqrt(B^2-4AC))/2A, so x=(-pi+-sqrt(pi^2-4sqrt(2)))/2=tan(angle).
    The two possible values of arctan(x) are -.4986 and -1.2032 in radians.
    Simply plug these values in for a and b in the second equation.
    Last edited by mr fantastic; April 10th 2009 at 09:23 PM. Reason: Added quote
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  3. #3
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    Quote Originally Posted by Media_Man View Post
    (1)
    You can use the quadratic formula to find a and b (alpha and beta).
    x=(-B+-sqrt(B^2-4AC))/2A, so x=(-pi+-sqrt(pi^2-4sqrt(2)))/2=tan(angle).
    The two possible values of arctan(x) are -.4986 and -1.2032 in radians.
    Simply plug these values in for a and b in the second equation.
    thanks media_Man , but we cant use calculator. and we must find exact value of A .
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  4. #4
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    No problem...

    but we cant use calculator. and we must find exact value of A
    No problem. Let a=(-pi+sqrt(pi^2-4sqrt(2)))/2 and b=(-pi-sqrt(pi^2-4sqrt(2)))/2. Then a+b=-pi. Verify this for yourself, and I think you can take it from there...
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  5. #5
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    Quote Originally Posted by linda2005 View Post
    [snip]
    2) let x,y,z > 0 , prove that : \frac{xy}{z} + \frac{yz}{x} + \frac{zx}{y} \geq x + y + z
    [snip]
    Pick three random positive numbers x,y,z.

    Define vector u=(xy,yz,xz) and vector v=(xz,xy,yz) in R^3

    Note: |u|=|v|, so let d=|u|=|v|

    Now we ask, which is bigger, u*v or u*u? (where * signifies vector product where appropriate)

    Since, u*v=|u||v|cos(a), for a the angle between them,
    u*v=d^2cos(a) and u*u=d^2cos(0)=d^2.

    Since all values are positive, u and v are confined to the first (+,+,+) quadrant, the maximum angle between them is 90*=pi/2 radians.

    Since cosine is a decreasing function on the interval [0,pi/2], cos(a)<cos(0), therefore u*u>u*v. You can also think of the fact that for two vectors of the same length, the dot product increases the closer they are together (the smaller the angle between them), so the maximal dot product for any vector is the dot product with itself.

    Substituting, this gives us: (xy)^2+(yz)^2+(xz)^2 > yzx^2+xzy^2+xyz^2. Divide by xyz on both sides and you have your exact answer.

    I KNOW THIS FEELS LIKE USING A CANON TO KILL A MOSQUITO, BUT THIS THE LEAST CONFUSING WAY I WAS ABLE TO VISUALIZE THE PROBLEM. I HOPE YOU ARE ABLE TO USE VECTOR GEOMETRY IN YOUR PROOF.
    Last edited by mr fantastic; April 10th 2009 at 09:24 PM. Reason: Added quote.
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  6. #6
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    Hello, Linda!

    1) Let \tan A and \tan B be roots of equation: x^2 + \pi x + \sqrt{2} \:=\: 0

    Evaluate: . X \;=\; \sin^2(A+B) + \pi\sin(A+B)\cos(A+B) + \sqrt{2}\cos^2(A+B)
    Since \tan A,\:\tan B are roots of: . x^2 + \pi x + \sqrt{2} \:=\:0

    . . we have: . \begin{array}{cccc}\tan A + \tan B &=& -\pi \\ \tan A\tan B &=& \sqrt{2} \end{array}

    Then: . \tan(A + B) \:=\:\frac{\tan A + \tan B}{1 - \tan A\tan B} \:=\:\frac{-\pi}{1-\sqrt{2}} \quad\Rightarrow\quad \tan(A+B) \:=\:\pi(\sqrt{2}+1)\;\;{\color{blue}[1]}


    We have: . X \;=\;\sin^2(A+B) + \pi\sin(A+B)\cos(A+B) + \sqrt{2}\cos^2(A+B)

    Divide by \cos^2(A+B)\!:\;\;\frac{X}{\cos^2(A+B)} \;=\;\frac{\sin^2(A+B)}{\cos^2(A+B)} + \frac{\pi\sin(A+B)\cos(A+B)}{\cos^2(A+B)} \;+  \frac{\sqrt{2}\cos^2(A+B)}{\cos^2(A+B)}

    . . . . X\sec^2(A+B) \;=\;\tan^2(A+B) + \pi\tan(A+B) + \sqrt{2}


    Substitute [1]:
    . . X\sec^2(A+B) \;=\;\bigg[\pi\left(\sqrt{2}+1\right)\bigg]^2 + \pi\bigg[\pi\left(\sqrt{2}+1\right)\bigg] + \sqrt{2}  \;=\;\pi^2\left(4+3\sqrt{2}\right) + \sqrt{2}

    . . . . . . . . . X \;=\;\frac{\pi^2\left(4+3\sqrt{2}\right) + \sqrt{2}}{\sec^2(A+B)}


    Since \sec^2\!\theta \:=\:\tan^2\!\theta + 1, we have:
    . . \sec^2(A+B) \;=\;\tan^2(A+B) + 1 \;=\;\bigg[\pi\left(\sqrt{2}+1\right)\bigg]^2 + 1 \;=\;\pi^2\left(3 + 2\sqrt{2}\right)+1


    Therefore: . X \;=\;\frac{\pi^2\left(4+3\sqrt{2}\right) + \sqrt{2}}{\pi^2\left(3+2\sqrt{2}\right) + 1}

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  7. #7
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    Incomplete

    Ah, yes. I see what I did wrong.

    Your proof is incomplete. Factor out a sqrt(2) from the numerator to give you:

    N=sqrt(2)(pi^2(2sqrt(2)+3)+1), which of course cancels with the denominator, leaving X=sqrt(2).
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  8. #8
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    Factor out a sqrt(2) from the numerator . . .

    Wow! . . . How did I miss that? . . . *blush*

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  9. #9
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    Quote Originally Posted by linda2005 View Post
    Hello, help please

    2) let x,y,z > 0 , prove that : \frac{xy}{z} + \frac{yz}{x} + \frac{zx}{y} \geq x + y + z
    expanding (a-b)^2 + (b-c)^2 + (c - a)^2 \geq 0 gives us a^2 + b^2 + c^2 \geq ab + bc + ac. now put a=xy, \ b = yz, \ c=xz and then divide by xyz.


    3) Find all function f: \mathbb{R} \to \mathbb{R} such that : f\left( xf(x) + f(y) \right ) = \left(f(x)\right)^2 +y. \ \ \ \ \ (*)
    this question is not bad! if in (*) we put x = 0 we'll get f(f(y))=(f(0))^2 + y. \ \ \ \ \ \ (**)

    1) f is onto: suppose z \in \mathbb{R} is given. put y=z - (f(0))^2. then by (**): \ f(f(y))=z.

    2) f(0)=0: by 1) there exists x with f(x)=0. then by (*): \ f(f(y))=y, for all y \in \mathbb{R}. let's call this result (***). now (**) gives us f(0)=0.

    3) \forall x \in \mathbb{R}: \ (f(x))^2 = x^2: if in (*) we put y = 0, we'll get f(xf(x))=(f(x))^2. again if in (*) we let y = 0 and replace x with f(x) and use (***) we'll get: f(xf(x))=x^2.

    4) f(x)=\pm x: by 3), for any x either f(x)=x or f(x)=-x. now suppose f(x)=x and f(y)=-y, for some non-zero x,y. then by (*): \ f(x^2 - y) = x^2 + y. on the other hand:

    f(x^2 - y)=\pm(x^2 - y), which gives us x=0 or  y=0. contradiction! therefore either \forall x \in \mathbb{R}: \ f(x)=x or \forall x \in \mathbb{R}: \ f(x)=-x.
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