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I can't help myself but there is some information out there for this interesting topic. Here is one such page:

http://numbers.computation.free.fr/Constants/Algorithms/fft.html

Here is the 7th one:

The key thing here is the Change of Base Formula:

Thus we should rewrite the function like this:

The last step was just to emphasize that 1/ln4 is just a constant multiplier.

Taking the derivative (Chain Rule) :

Here is the 4th one:

There is a really useful formula that can help with this. Let's derive it:

First, there is a crucial result that for all functions f:

So, we can write:

This is just e raised to a constant times x.

Taking the derivative:

But now this can be re-written as:

In general, it's better to just remember the formula:

And notice this still holds if b = e:

Here is the 3rd one:

The key to this is the chain rule, and knowing that e^x has the special property:

Thus:

The key things to know are: that you can write this as a sum of two fractions, that a negative power means *1 over the number raised to the positive power*, and the rules of adding and subtracting exponents.

Here is how it works out:

You need to solve the equation: "e raised to some power is 0.7165"

ie, solve:

To do that you need to take the natural logarithm of both sides:

m and n are any integers in this case.

But we could say they are both real numbers and the rule still applies.

The point is, x^0 = 1 is consistent with this rule and all the other rules of exponents. If it was defined as anything else, there would be inconsistency!

Because of the rule:

If n = m, then x^n/x^m will be 1... so x^(n - m) = x^0 must be 1 also.

You could also do this:

Now substitute this into the second part:

This has the advantage of only depending on y.

Did you round off a lot of the numbers in between steps? I get the answer as 1 exactly...

Thanks.

Two lessons for me: thinking more about Latex and less about algebra is bad - and: never assume the work so far is correct !

Well you can write that as :

Or:

(because

)You can write it as any of the above... I don't think there is a way to make it any simpler than those.

So, you first take the sin, and then square the result:

But are you supposed to get an exact value for this ? Or is a decimal ok?

You must apply *twice* a rule developed by a certain French mathematician.

But All_is_number's interpretation seems most natural, plus it doesn't give non-integer answers.

Yes you can do your question like that too -- except you can see the slope of your line is 2 already, without re-writing it. The key thing to know is: parallel lines have the exact same slope.

So your parallel line's equation will be y = 2x + b ... now you just need to find b. But you know a point on the line, so just substitute the point in this equation!

Oh right!

So the function actually reduces to:

Well we are trying to find

so let's restrict the values of x we are looking at to:

x is in that interval if it is approaching 2, and f(x) = -1 on that interval; it never reaches 2. So when evaluating the limit, f(x) = -1, and the limit is still -1 ... right?

Ok here is another way, no Fourier:

We need to be aware of this formula:

therefore:

And so:

It is evident that this formula always gives -1.

Eg for 2.1 , the floor is 2, the ceiling is 3, and we are always taking the floor - ceiling, which is always going to be -1.

Thus the function simply reduces to f(x) = -1 for all x.

This makes the proof of the limit trivial.

Oh yeah, f(2). haha thanks