All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Trig meets Logs (Posted on 2006-06-14)
Prove the equation cos(x) = log(2*pi)x has exactly 3 real solutions if x is measured in radians.

What must the base of the logarithm be reduced to to give the equation exactly two real solutions?

 No Solution Yet Submitted by Jer Rating: 1.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 non-rigorous "proof" for part 1; numerical computer solution for part 2 Comment 1 of 1

For x <= 0, the logarithm of x does not exist as a real number.

From x = 0 to x = pi, cos(x) is monotonically decreasing from 1 to -1, while log2pi(x) is monotonically increasing, from negative infinity to approximately 0.6228544372044725, so in this interval the two curves cross once.

In the range pi < x < 3*pi, cos(x) goes up monotonically once and down monotonically once, reaching a maximum at 2*pi, while log2pi(x) is increasing monotonically. At x=6.2, cos(x) = .9965420970232175 while log2pi(x) = .9927482775633427 so that the former is greater, while at the end points of this interval, log2pi(x) exceeds cos(x).  This does not actually prove that there are only 2 points of intersection of the curves in this range of x values, and this "proof" could use some strengthening here. A graph of the curves looks convincing though.

For x>=3*pi, cos(x) never exceeds 1, while log2pi(x) always exceeds 1.22 for these x values.

Turning to part 2:

In order for there to be only 2 real solutions, there should be only one solution between pi and 3*pi, so the  cosine curve should be tangent to the log curve.  Thus they should have the same slope.

The slope of cos(x) is -sin(x).

The slope of log2pi(x) is 1/(x*ln(2*pi))

We want -sin(x) = 1/(x*ln(2*pi))

or x = -1/(sin(x) * ln(2*pi))

The following spreadsheet was set up to solve this, initially with a guessed value of x in A2:

`     A                  B                        C         D         E1    x                  -1/(sinx ln(2pi))2    6.19524561970512   =-1/(SIN(A2)*LN(2*PI())) =A2-B23    =COS(A2)           =LN(A2)                  =B3/A3    =EXP(C3)  =LN(A2)/LN(D3)`

The above value in A2 is the result of using Excel's solver to set C2 to zero by varying A2.

The figures resulting are:

`x          -1/(sinx ln(2pi))6.19524562  6.19524562      4.41069E-110.996135797 1.823782162      1.830856965 6.239231174 0.996135797`

where C3 was set to see what conversion factor was needed to bring the natural log value down to the value of the cosine function.  Taking the natural antilog of this is done in D3, giving the necessary base for the logarithms to produce this point of tangency: base 6.239231174.

Returning to part 1:

When pi < x < 3*pi, the slope of the original log function varies between 0.5110502023754473 and 0.086926022918518, while the slope of the cosine function is for a large part larger than this. The fact that it is not always larger than this again prevents this from being a rigorous proof.  But again, a look at the graphs of the functions is convincing in an informal way.

Edited on June 14, 2006, 1:09 pm
 Posted by Charlie on 2006-06-14 13:08:05

 Search: Search body:
Forums (0)