Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Solving trig/algebric equations

  1. Sep 16, 2010 #1
    I was finding areas between random lines to pass time when i encountered an equation i could not solve....

    7sin(x)=x+1

    Do i have to replace sin(x) with the taylor series equivalent
    29cb648e96291f52707c7225630c1a17.png

    Or is there another way to solve it ?

    Thank you.
     
  2. jcsd
  3. Sep 16, 2010 #2

    Mentallic

    User Avatar
    Homework Helper

    You can only get an analytical solution at best, i.e. a good approximation for x.
     
  4. Sep 16, 2010 #3
    Sure, I'm looking for a good approximation... Would it involve plotting the function f(x)=7sin(x)-x-1 on a graph and picking points on the x axis or is there a mathematical method to solve something like this...

    How would you solve it ?
     
  5. Sep 16, 2010 #4

    Mentallic

    User Avatar
    Homework Helper

    Oh ok as long as we're clear we can't solve this algebraically :smile:

    Yes you would use the function f(x)=7sin(x)-x-1 and to find a good approximation for this function, I'd probably use "[URL [Broken] method[/URL].
     
    Last edited by a moderator: May 4, 2017
  6. Sep 16, 2010 #5
    Unfortunately I don't really know what to do to solve it.... Could you solve this specific example using Newton's method ?
    If it is too long, then perhaps you could get me started in the right direction (the Wikipedia article fried my brain, I'm not even in college...)

    Thank you.
     
  7. Sep 16, 2010 #6

    Mentallic

    User Avatar
    Homework Helper

    Yes it definitely works for this problem.

    It is defined as follows:

    [tex]x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}[/tex]

    [itex]x_n[/itex] is your first estimated solution, and then [itex]x_{n+1}[/itex] will be your next more closely approximated solution (as long as you use a value of x that is close enough to the actual value).

    Now, [tex]f(x)=7sinx-x-1[/tex] has 5 real roots so depending on which root you're looking for, choose an x that is quite close to it to begin with. Keep applying Newton's method and it should give you a very good approximation pretty quickly.
     
  8. Sep 17, 2010 #7
    Wow ! That was fast... Thanks a lot !
    I have one more question though...

    I used Newton's method for some random polynomials like x3+4x2-3x+4=0 (easily done on a calculator, but this was more fun) keeping my xn as 0....
    Finding my first value (x=-4.8) I used long division to divide my expression by (x+4.8) and then used Newton's method for the remaining polynomial, thus finding all the 3 factors of the expression...

    However, I am confused when it comes to my previous function...
    How do i get all 5 values of x for f(x)=7sin(x)-x-1 ?
    Do I use Newton's method to find the first factor and then change my f(x) to:
    f(x)=(7sin(x)-x-1)/(x-a) [where a is the first factor i get]
    and then use the method again and again...

    I have not tried this yet because differentiating it will become progressively more painful...
    Will this work and/or is there any other way to get all 5 factors ?

    Thanks
     
  9. Sep 17, 2010 #8

    Mentallic

    User Avatar
    Homework Helper

    You couldn't have found the other factors of that cubic... you divided by an approximation for the real root. [tex]x\approx -4.8[/tex] but really, the exact value of x is irrational and expressing it is complicated.
    For example, say we have the function y=x-1, if we divide that by x-(1+0.001) then we now have some rational function which still has a root of 1 and instead now has a vertical asymptote at x=1+0.001.

    Since your equation is mixed with trig and algebra, we can't solve the roots algebraically so the best we can do is get a good approximation, which means we can't divide by the root in order to simplify the expression.

    The best way would probably be to use a graphing calculator or a carefully drawn diagram of the functions y=7sin(x) and y=x+1 to give yourself a good starting point to find where the roots lie approximately, at which point you can use newton's method again.
     
  10. Sep 17, 2010 #9
    Darn, that's too bad, I'm never too good with graphic approximations...

    Is there a threshold range for which Newton's method will return a particular value of x?
    That is, if f(x)=sin(x), what is the range of values for xn for which Newton's method will give me x=0 and not jump to x=+[tex]\pi[/tex] or x=-[tex]\pi[/tex]

    Is there a derivation for this range based on the function f(x) ?

    Thanks
     
  11. Sep 19, 2010 #10
    Mentallic? have any idea how close the first approximation needs to be to the actual value to get the correct one ?
     
  12. Sep 19, 2010 #11

    Mentallic

    User Avatar
    Homework Helper

    If you haven't got a graphing calculator handy and just have no chance of getting a decent approximation by hand-drawing it, you can use the intermediate value theorem. This is pretty simple. Say you have y=x2-2 and want to find where y=0, thus x=sqrt(2), by your graph you estimate it is between 1 and 2, checking to make sure, you find the value of y at x=1, f(1) and notice that it is less than zero, so you check f(2) and notice it is more than zero. So this means the root is somewhere in between x=1 and x=2, take x=1.5, it is slightly more, so take x=1.25, slightly less... so take half way between 1.25 and 1.5.. etc. It's a slow process but it will give you a good head start to kick off your newton's method approximation.

    I don't believe there is. The way newton's method works is that with your first approximation, it finds where the function actually is at that x value, and then calculates the x value of the tangent line that cuts the x-axis. This is the next approximation and it just keeps doing this over and over.
    Say for sin(x), if you take [itex]x=\pi/4\approx0.78[/itex] as your first approximation, then newton's will bring back the value [itex]x=\pi/4-1\approx -0.21[/itex] and it is clear that since the second value brought back something closer to the true root, it will converge to x=0. If you took say [itex]x=\pi/2-0.001[/itex] then the tangent line from this point is nearly flat so it will reach a very long distance before touching the x-axis again to give the next approximation. Your next value would be [itex]x\approx -998[/itex] so obviously you're screwed here. But you can even take x=4.5 and return x=-0.1 so yeah, just keeping the rule of thumb that you need a decent first approximation will come in handy. It really depends on the function too,

    say, you want to find the root x=0 of [tex]y=x^{1/3}[/tex] then applying newton's method: [tex]x_{n+1}=x_n-\frac{x_n^{1/3}}{\frac{1}{3}x_n^{-2/3}}=x_n-3x_n=-2x_n[/tex] So whatever first value you use, the next will always double and become negative of it, thus never converging to 0.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Solving trig/algebric equations
  1. Trig equations (Replies: 14)

  2. Trig Equation (Replies: 12)

  3. Trig Equations (Replies: 8)

Loading...