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#1 Re: Help Me ! » fast fourier transform » 2007-07-21 07:35:25

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

#2 Re: Help Me ! » Derivatives » 2006-11-30 06:30:42

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) :

#3 Re: Help Me ! » Derivatives » 2006-11-30 06:21:49

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:

#4 Re: Help Me ! » Derivatives » 2006-11-30 06:04:45

Here is the 3rd one:

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


#5 Re: Help Me ! » simplify plz help » 2006-11-29 09:12:42

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:

#6 Re: Help Me ! » correct » 2006-11-09 14:47:50

I thought that whenever the square root sign is written, as in a function definition for example, it means "extract the positive root"... at least it is like that in all the textbooks i've seen. I have never seen any instance where a function definition involving a square root operator was supposed to mean both possible roots.

#7 Re: Help Me ! » To the power of 0 » 2006-11-06 15:47:08

re: the difference between "undefined" and "indeterminate" -- i think undefined means strictly "there is no meaningful way to define the symbol or operation", so we leave it undefined, 'illegal'. The term 'Indeterminate' is applied only to limit situations, ie, when we have a function like (x-1)/(x^2 - 1) and x approaches 1, the function's value approaches an indeterminate value -- not undefined, but *we are not able to determine the value being approached*.

#8 Re: Help Me ! » raised power » 2006-11-06 07:19:44

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:

#9 Re: Help Me ! » To the power of 0 » 2006-11-04 03:26:44

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!

#10 Re: Help Me ! » To the power of 0 » 2006-11-04 02:36:56

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.

#11 Re: Help Me ! » help me please » 2006-11-01 17:48:05

You could also do this:

Now substitute this into the second part:

This has the advantage of only depending on y.

#12 Re: Help Me ! » sin cos » 2006-11-01 13:01:35

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

#13 Re: Help Me ! » Exact value for an angle. » 2006-11-01 12:55:35

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

#14 Re: Help Me ! » Exact value for an angle. » 2006-11-01 11:05:41

Well you can write that as :




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

#15 Re: Help Me ! » sin cos » 2006-11-01 10:56:59

That happens because your calculator is in Degree Mode. You need it in Radian mode, because angles like 7pi/6 are in Radians.

#16 Re: Help Me ! » sin cos » 2006-11-01 10:45:42


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?

#17 Re: Help Me ! » I just want a hint (Limits & Calculus) » 2006-11-01 10:38:58

Yep, you can keep applying it any number of times, as long as the expression is an indeterminate form !

#18 Re: Help Me ! » I just want a hint (Limits & Calculus) » 2006-11-01 10:03:47

I don't know how much of a hint you want.. so here is a vague one:
You must apply *twice* a rule developed by a certain French mathematician. smile

#19 Re: Help Me ! » Measuring perimeters » 2006-10-31 09:57:44

But what about a cone -- the distance around the tip is 0 (from the side view), the distance around the base is something else, the distance around a side is different .. there are an infinite number of choices!

#20 Re: Help Me ! » Need answer for this problem » 2006-10-31 05:03:04

That was my very first interpretation actually, and I got 436.667 minutes to 6PM.
But All_is_number's interpretation seems most natural, plus it doesn't give non-integer answers.

#21 Re: Help Me ! » correct » 2006-10-31 04:58:18

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!

#22 Re: Help Me ! » problem my son brought home from school » 2006-10-31 04:51:48

Yeah, it doesn't make much sense to talk about the perimiter of 3D objects. However, an object like a ruler or a thin notebook with a tiny width can be thought of as 2D and one can measure their perimeter.

#23 Re: Help Me ! » Largest integer function ! » 2006-10-29 13:02:41

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?

#24 Re: Help Me ! » Largest integer function ! » 2006-10-29 04:59:38

Ok here is another way, no Fourier:

We need to be aware of this formula:


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. smile

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