Results 1 to 8 of 8

Thread: linear function

  1. #1
    Member
    Joined
    Dec 2017
    From
    Tel Aviv
    Posts
    142

    linear function

    Can I consider that cos(a)+cos(b) != cos(a+b)
    because it not a linear function?
    In the definition of linear function a & b are vectors.
    I know from physics that vector is contain size and angle.
    So I can say because a & b are not vectors and the expression isn't linear function, this is the reason that the expression isn't valid?

    and: When the expression is consider valid?

    the expression:
    cos(a)+cos(b)=cos(a + b).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2010
    Posts
    3,717
    Thanks
    1510

    Re: linear function

    See Wolframalpha

    It is an infinite number of ellipses.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    22,073
    Thanks
    2977
    Awards
    1

    Re: linear function

    Quote Originally Posted by policer View Post
    Can I consider that cos(a)+cos(b) != cos(a+b)
    because it not a linear function?
    In the definition of linear function a & b are vectors.
    I know from physics that vector is contain size and angle.
    So I can say because a & b are not vectors and the expression isn't linear function, this is the reason that the expression isn't valid?

    and: When the expression is consider valid?

    the expression:
    cos(a)+cos(b)=cos(a + b).
    Please read carefully this webpage.
    You should know that $\large{\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)}$

    One of the fundamental properties of linear functions is: $\large{f(a+b)=f(a)+f(b).}$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Dec 2017
    From
    Tel Aviv
    Posts
    142

    Re: linear function

    Quote Originally Posted by SlipEternal View Post
    See Wolframalpha

    It is an infinite number of ellipses.
    they are look like ellipses but they are not actually ellipses (Right?).
    What the reason of that? [of looking like an ellipses (perfect ellipses figure...)?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Dec 2017
    From
    Tel Aviv
    Posts
    142

    Re: linear function

    Also:
    What change I need to make in the expression to make the graph look like an ellipse or ellipses (...prefect ellipse or perfect ellipses)?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Nov 2010
    Posts
    3,717
    Thanks
    1510

    Re: linear function

    It does not display it, but it also will generate a "grid". Basically, the lines $x=2\pi m$ and $y = 2\pi n$ for $m,n \in \mathbb{Z}$ will work. I am not sure how close or far from a perfect ellipse it may be, nor do I really want to spend time thinking about it.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Nov 2010
    Posts
    3,717
    Thanks
    1510

    Re: linear function

    I was wrong on the gridlines. Never mind that. Try rotations 45 degrees to turn "unslant" it. Then shift it by $\pi$ to the left and down. Then find when $x=0$ and when $y=0$. The four points closest to the origin will be correct. Find the ellipse with major an minor axes equal to what you find. Show that you do not have an ellipse by plugging in other points. I'm not sure how to answer "how close is it to an ellipse"?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Nov 2010
    Posts
    3,717
    Thanks
    1510

    Re: linear function

    First, we shift it by $\pi$ to the left and down:

    $$\cos(x+y+2\pi) = \cos(x+\pi) +cos(y+\pi)$$

    This gives us the translated equation:

    $$\cos(x+y) = -\cos x - \cos y$$

    It should be obvious that if $(a,b)$ is a solution to the equation, then so too is $(b,a)$. This means it is symmetric across the line $y=x$. So, we can use the change of variables:

    $$x' = x\cos \dfrac{\pi}{4}+y\sin \dfrac{\pi}{4} = \dfrac{x+y}{\sqrt{2}}$$

    $$y' = -x\sin \dfrac{\pi}{4}+y\cos \dfrac{\pi}{4} = \dfrac{y-x}{\sqrt{2}}$$

    Solving for $x,y$, we get:

    $$x = \dfrac{x'-y'}{\sqrt{2}}$$

    $$y = \dfrac{x'+y'}{\sqrt{2}}$$

    Plugging in, we have:

    $$\cos(x'\sqrt{2}) = -\cos\left(\dfrac{x'-y'}{\sqrt{2}}\right)-\cos \left(\dfrac{x'+y'}{\sqrt{2}}\right)$$

    If you graph this, you will see it is a bunch of ellipse-like shapes with one centered at the origin. That is the one we will play around with to see how ellipse-like it really is.

    Plugging in $x'=0$ gives $y' = \pm \dfrac{2\sqrt{2}\pi}{3}$.

    Plugging in $y'=0$ is a bit more difficult to solve, so I just plugged it into Wolframalpha and got:

    $$x' = \pm 2\sqrt{2}\tan^{-1}\left(\sqrt{2\sqrt{3}-3}\right)$$

    Suppose this were an ellipse. Then, can find the lengths of the major and minor semi-axes, and write the formula. When you plug in points, you will find that they do not match, so the figure cannot be an ellipse. Figuring out how close or how far it is from an ellipse is far more difficult.

    Plugging in, we can get the area of the shape we create with the function: Wolframalpha. It comes to about 15.7374.

    Next, we can get the area of the ellipse with points we calculated:

    $$\pi \dfrac{2\sqrt{2}\pi}{3}\cdot 2\sqrt{2}\cdot \arctan\left(\sqrt{2\sqrt{3}-3}\right) \approx 15.7395$$

    So, it is really close to an ellipse.
    Last edited by SlipEternal; Aug 24th 2018 at 12:24 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Linear Function
    Posted in the Number Theory Forum
    Replies: 10
    Last Post: Jun 8th 2014, 09:43 AM
  2. Quadratic Function + linear function
    Posted in the Algebra Forum
    Replies: 4
    Last Post: Dec 20th 2011, 06:46 PM
  3. linear function
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Oct 18th 2009, 03:38 PM
  4. Linear Function
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Dec 22nd 2008, 06:32 AM
  5. Linear Function
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: May 2nd 2007, 08:06 AM

/mathhelpforum @mathhelpforum