What is the integral of 0 for infinite iterations of 3?

[mathelord]

first the integral of 0 should be a constant rather any constant,which makes the integral of 0 to be 1=2=3=4...
does this equate all numbers or am i gettin this mixed up.
well that is not want i intend sending,find the derivative of 3^3^3^3^3^3......^x,and then tell me what the integral of 0 really is.I rest my case

 

[Icebreaker]

You're not making sense. The antiderivative of 0 is C, that is because information is lost when deriving. The derivative of any constant is 0. Where does "1=2=3=4" come from?

What's the difference between "a constant" and "any constant"?

Also, the definite integral of 0 on any interval is 0.

 

[Cyrus]

If you have an integral of zero, and you evaluate it, you will get 0+c. C is "a constant" But you can't just pick and choose a value for c. In order to find the value of c, you need the value at a point on the function. This will determine the PARTICULAR value of c you have to use. So its not literrally "any constant", it depends on what value the curve has. 1=2=3=4 does not happen, I am not sure how you came to that conclusion, but I think it was via an error in an assumption along the way.

 

[Alkatran]

Maybe it's time mathematicians start putting up massive billboards:

'MANY TO ONE' FUNCTIONS DO NOT SHOW THAT 1 = 0. 'ONE TO MANY' FUNCTIONS DO NOT SHOW THAT 1 = 0. 1 DOES NOT EQUAL 0.

 

[HallsofIvy]

I think I saw one of those on I95!

 

[mathwonk]

you guys do not seem to realize that the word "integral" does NOT mean antiderivative. the integral of zero, over any interval at all, is definitely just zero.

 

[SteveRives]

you guys do not seem to realize that the word "integral" does NOT mean antiderivative. the integral of zero, over any interval at all, is definitely just zero.

Yes! Exactly! And with this we are free to muse about the integral on its own terms -- it is its own thing. Which allosws us to ask the question: what is zero growing into as we add zero to it?

Such a question is a geometric-like way of thinking of the integral. And, as such, no one needs to think about antiderivatives to realize that such a thing is "definitely just zero."

 

[Cyrus]

What ever happened to the constant of integration?

 

[Cyrus]

I can't read your text because its not long enough rach, sorry. I always thought evaluating the integral was to do the antiderivative to it. Could you explain the difference please.

 

[quasar987]

The integral evaluated from a to b is 0, but the improper integral, i.e. the anti-derivative, is C.

 

[rachmaninoff]

Specifically, the FTC tells us that \int_a^b 0 dx= F(b)-F(a)=C-C=0. An antiderivative is just a function, but an integral is a number.

 

[MalleusScientiarum]

You don't even need the fundamental theorem of calculus. You can start from the Riemann-Steltjes definition of the integral and prove that any finite sum of zeros is zero.

 

[quasar987]

Of course, but it's nice to see how the two "views" concord with each other. I.e. how the FTC "makes it work".

 

[rachmaninoff]

You don't even need the fundamental theorem of calculus. You can start from the Riemann-Steltjes definition of the integral and prove that any finite sum of zeros is zero.

Hey, I was just trying to make things easier to understand. If you're looking for the most general result, then might as well point out that a zero function (real or complex) has Lebesgue integral zero over any measurable set in any measure space.

 

[Cyrus]

Ah, yes of course. Thanks rachmaninoff. From the wording of the original question, I thought it was in refrence to doing the antiderivative.

 

[mathelord]

so what does the derivative of the function 3^3^3^3^3^3^3^3^......^x gives us

 

[Icebreaker]

\frac{d}{dx} 3^x = 3^x\ln 3

\frac{d}{dx} 3^{3^x} = 3^x \ln 3 * 3^{3^x} \ln 3

You can do the rest. However, I don't see how this has anything to do with the integral of zero or how 1=2=3=4.

 

[hypermorphism]

so what does the derivative of the function 3^3^3^3^3^3^3^3^......^x gives us

Let the function T(a) = 3^a. The derivative of this function with respect to a is 3^a \ln(3). Thus you have the function T(T(...(T(x))...)) where T is composed n times with itself. Let 3_i(x) be the power tower of 3 to order i where the ith position is replaced with the variable x, and 3_i be the power tower of order i. The derivative of the given function is then \frac{d}{dx}T(T(...(T(x))...)) = T'(T(...(T(x))...))*T'(...(T(x))...)*...*T'(x)
= \prod_{i=1}^n \ln(3_{i-1})*3_i(x)
Nasty looking thing.

 

[mathelord]

i don't understand,ice breaker i do not buy your idea

 

[LeonhardEuler]

Do you mean (3^3^3^3...^3)^x or 3^(3^(^3(...3^(x)))))))... . If you mean the first case then this is just(3^3^3^3^3...^3)^x = e^(ln((3^3^3^3^3...)^x))= e^(x*ln(3^3^3^3^3...^3)) and the derivative is therefore ln(3^3^3^3^3...)e^(ln((3^3^3^3^3...)^x)) = ln(3^3^3^3^3...)(3^3^3^3...^3)^x. In the second case it is like hypermorphism suggested.

 

[Icebreaker]

What does this have to do with your "case" that a number is equal to any other?

 

[mathelord]

no it has nothing to do with that,all i asked mainly was to find the derivative of the function.3 is raised to 3 which is raised to another 3 and so on eventually the last three is raised to x.hope u get it this time

 

[Icebreaker]

What, I didn't have it before? All you have to do is apply the chain rule.

 

[rachmaninoff]

How many 3's are there?

 

[mathelord]

infinite 3ssss
and the last three carries the x.what i mean is the latter of leonhardeuler expression.
and i do not understand wat hypermorphism did

 

[Icebreaker]

You can't have infinite 3's.

 

[mathelord]

3^3^3^3^3^3^3^3...
and the last 3 carries the x

 

[EL]

But if there's an infinite number of 3's, then there is no "last one".