Math Is Fun Forum

For maximum reliability:

1. z is equal to x to the power of open bracket y plus open bracket 2 divided by x close bracket close bracket.

Actual attempt:

1. z is equal to x raised to the sum of twice the inverse of x and y.

2. z is equal to x raised to the quotient of y + 2 and x.

3. z is equal to twice the inverse of x, plus the yth power of x.

Hmm, right! So i'm confused again.

Using the idea of evaluating the most deeply nested function first, if we for example in the case of the function i mentioned write:

Then the x^2 must be evaluated first because it then is an argument to the exponential function!

Thanks. I'm not sure why I was thinking it is actually the other way around.. for example when evaluating:

It might seem natural to square the x first, but then the curve would be all wrong!

Right, oops.

So it doesn't look like there is a way to do this by doing nothing but moving only 1 stick on only the left side, unless we are allowed to replace it with an operator or using it to cross out the equals sign or something.

He means the sinc function sinc(x).

http://mathworld.wolfram.com/SincFunction.html

t is the independent variable, which would be time in this case, on the horizontal axis.

e is the base of the natural logarithm, and is equal to about 2.718281828

e^t is just the standard exponential function that is used to model all kinds of exponential growth in science.

here is an example of such a curve (the top one in this picture):

http://cognitrn.psych.indiana.edu/busey/WWWPubs/PsychRev/Image2.gif

k, which I called a rate constant, is a number that determines how sharply the curve rises at the beginning.

If k = 100 for example, it will rise very sharply, if k = 1, the rise is moderate.

I would suggest plotting a few of these with graph plotting software, like Winplot (which is free and easy to use).

Therefore:

It seems like the paper proves the *existence of a solution*, more specifically the existence of an 'immortal smooth solution' (whatever that is)... is this enough for the prize?

Well I think that's all you can do. By taking out the common factor 2x^(3n + 2), you get:

I think it's a little difficult to see how to get 3x^(n-1) just like that, so perhaps its easier to do it this way:

Now here is the tricky part: if we factor the other 2x, we need to take out an x in the 3x^n term.

We do that by writing:

Because x multiplied by 3x^(n - 1) gives us the original 3x^n.

If you mean:

Then what you need to do is expand it out and use the laws of exponents:

It might be possible to simplify this more but I can't think of how at the moment.

I would suggest carrying out polynomial long division, and then what you obtain from that would be a proper fraction plus another term.

Then the proper fraction has known cases of partial fraction decomposition, which is all straighforward.

As John said, an exponential approaching 100% might work nicely.

It has an asymptote at y = 1.

Such a curve would be:

Where k would be a rate constant.