Solve for x in log(x)*sin(x) ???

[MikkyThomeon]

Really don't know how to get this one. Its been a while since I have been in maths stuff. Your help would be appreciated

[NoteMe]

Not sure if I got this one...

doesn't it have to be equal to something to be possible to find X?


log(x)*sin(x) = ?

[MikkyThomeon]

f(x) = log(x)*sin(x)

so if y = f(x), then what would the value of x be given a particular value of y...

This is a sine wave that grows in a logarithmic fashion....

[alkatran]

You can't "solve" a function. You can only graph it. You said it yourself, it's a sine wave that grows logarythmacilly (ok, the spelling went off at the end there(

[MikkyThomeon]

With all respect due, a function or any product or sum thereof can be represented on a 2D set of axes. Just like in the graph of

y = m*x + c (straight line)

a particular value of y exists for a particular value of x. You can also reverse the situation and say that a particular value of x exists for a particular value of y.

Hence it is possible to create an expression of y = log(x)*sin(x) in terms of x. f(y) = ???

I am saying that, if y = 5, then what is the value of x at that point?

If you break things up a bit and think about what this waveform would look like for a single cycle, it would be a single sine that has an increasing amplitude that increases in a logarithmic fashion. It must surely be possible to find an expression in terms of x that represents y...

[TheManWhoCan]

No such expression exists.

You'd run into the same problem if you tried to solve

y = f(x) = x - sin(x)

for x...

[bugzpodder]

your best bet is try some numerical technique... assuming this is used in a computer setting

[alkatran]

This is the sortof thing you need to solve with a loop (since there are infinite answers at any one point) with a lower limit that increases as you go. Basicly, your equation is probably going to cover a page...

[Darkwraith]

You can't "solve" a function. You can only graph it. You said it yourself, it's a sine wave that grows logarythmacilly (ok, the spelling went off at the end there(

You could solve and equation for x. For example, if we had the equation:

y = m*x+b

we could solve this for x, giving us:

(y - b) / m = x

I tried running the equation through my calculator, and I could not get a solution.

[alkatran]

I said in my second post that it would be a "weird" formula.

For example, when I put Solve(Y = sin(x), x) into my calculator (isolate x), it doesn't return arcsin(Y), it gives me back something like:

x = (-180*(pi*arcsin(y)/180-(2*@n28+1)*pi))/Pi and y <= 1 and y >= -1 or (same basic thing)

The key here is the (2*@n28+1) where (I think) @n represents all whole numbers.

[Darkwraith]

Ok... I thought that you meant that we could not solve any equation for x.

I did not say that my calculator gave me a weird formula for log(x)*sin(x). I said that my calculator gave me nothing back (as in it cannot solve for x.)

[alkatran]

Mine can't either

All it gives back is:

ln(x)*sin(x) = ln(10)*y

Wow, it converted log to ln, then moved the division to the other side.

[jemidiah]

If you knew the number of decimal places you need (and the range of values, though modulus would take care of that), you could use a finite number of terms in the Taylor series for sine and attempt to solve that way. The calculator might be able to help more then.

[MikkyThomeon]

Thanks for your responses. It seems like numerical methods are the way then which is OK if a machine is going to do the calculations. It seems that our "normal" mathematics has a few inabilites when trying to derive solutions such as these

[krtxmrtz]

As some folks have pointed out it's not possible to reverse this equation by analytic methods, therefore numerical methods are the only way (so what?)
So you must find the zero(s) of
log(x)*sin(x) - y = 0
for a particular value of y.
Numerical methods usually rely on choosing an approppriate initial value for x and repeatedly applying a recursive algorithm whereby new values are found for x that may or may not converge. One thing to bear in mind is, because sin(x) has infinite zeors, so will your equation, so that depending on the particular choice of the initial values you'll get different solutions.

For this equation I would try a scheme like this:

1) Split your equation into 2, for instance:
f(x) = log(x) - y
g(x) = sin(x)
2) Select an initial value for x, (call it x₀)
3) Substitute x₀ into g(x), i.e., find g(x₀) = sin(x₀)
4) Use this value as f(x) and calculate x₂ such that
f(x₁) = log(x₂) - y
5) Calculate g(x₂)
6) If abs(x₂ - x₁) < epsilon you're done (epsilon is the accuracy you want to achieve). Otherwise go to step 4 and so on.