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This is another problem that I need help on.
1. Prove that there exists an infinitely differentiable function.
I was thinking I should first find a function that is infinitely differentiable then prove that it is infinitely differentiable. So I was thinking about show that the function f(x)=x is infinitely differentiable which I beleive it is. So can anyone help me with this.
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What do you mean by "infinitely differentiable"? Do you mean that its derivative exists for any point on the real number line? if so, then I would think f(x)=x is fine, and I'm sure you can prove it with epsilons and deltas.
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My guess would be that we'd be looking for a function that can be differentiated any number of times (take the first derivative, then take the derivative of that to get the second, and so on). Now, the question is whether the result needs to be nonzero. Otherwise, f(x)=x would work just fine, although the derivatives get pretty boring after the first one.Originally Posted by Normandy6644
If the instructor is looking for one that has some nonzero value no matter how many times you differentiate it, you need a function which doesn't get simpler with each one like the algebraics do. How about a function which when differentiated gives itself, or perhaps a pair of functions which have each other as their derivatives? I can think of a couple examples of these, can you?
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Combinations of those awesome functions are always fun, too.
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NM My mistake, he does want us to find a function in which the dervative can be taken an infinite number of time. So I used f(x)=e^x, and i was able to prove it implicitly. Thank you for the help.
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Yep e^x is a good one.
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This is my proof.
f(x) is infinitly differentiable if and only if f(x) = f^n(x).
Let f(x) = e^x, and suppose that f^n(x)= e^x
i) f^1(x)= e^x ( holds when n=1)
ii) Suppose f^n(x) = e^x, then D(f^n(x)) = D(e^x)
Thus f^(n+1)(x)= e^x. ( holds if n is true then n+1 is true)
Therefore f(x)=e^x is an infinitly differentiable function.
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Yep, that works nicely. Gotta love induction!
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Also does anyone know the difference between PMI and PCI induction because I dont.
PMI Principle of Mathematical Induction
PCI Principle of Complete Induction
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What about sin(x)? And cos(x)? They're infinitely differentiable yet the derivative of sin(x) does not equal sin(x). Of course the fourth derivative of sin(x) equals sin(x), but I don't think the most general infinitely differentiable function has to satisfy a recurrence relation like this. I haven't tried, but I am fairly certain that Bessel functions are infinitely differentiable too.
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I was thinking about sine and cosine too. I don't see anything wrong with them being infinitely differentiable. I don't know enough about Bessel functions to say anything about them though. I bet there are probably some conditions an infinitely differentiable function would satisfy, but I don't know what they are.
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I was gunning for e^ix, myself.
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Your right, then my proof should read "if" instead of "if and only if".
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This isn't perfect (see below), but to prove a function is infinitely (continuously) differentiable, say in some open interval, it is sufficient to prove that it has a convergent power series in that interval (the power series is then the Taylor series, and you can read off the derivatives). In particular, any polynomial works, as does the exponential, and sine and cosine.
Interestingly enough, though, there are infinitely (continuously) differentiable functions whose Taylor series *don't* converge to the original function everywhere. For an example, take the function
f(x) = exp(-1/x^2) for x > 0 and f(x) = 0 for x <= 0. One can show that this function is infinitely continuously differentiable, but if you try to do a Taylor series centered at x = 0 for a little neighborhood around zero, you just get the zero function, and our function f is nonzero for positive x.
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Sorry for posting twice in a row; didn't see this at first.
If I recall correctly, complete induction is when, while trying to prove something is true for n, you assume it it true for *all integers* less than n, whereas in mathematical induction you simply assume it is true for the integer immediately less than n. They both work by starting somewhere and working your way up, so they are equally valid (so you still have to show it works for n=1 or wherever you're starting). Typically, you'll see complete induction used when you reduce the case for n to cases for integers less than n, but not necessarily immediately preceding n. For example, to prove that every integer greater than one has a prime factorization, you simply
1) note that 2 is prime, hence has a prime factorization (so we have our starting point)
2) Let n be any integer greater than 2; assume every integer less than n has a prime factorization. Then if n is prime, we're done. If not, n can be written as the product of two numbers strictly less than n, both of which have prime factorizations by our assumption, so n does.
EDIT - hehe maybe I should be a little more clear and say every number is prime or can be written as a product of primesops: So for the above argument, this is what is meant by 'prime factorization'
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Unless you consider taking the derivative of 0 as part of "infinitely differentiable" I don't think any polynomial works at all. In fact, no polynomial should work since the derivative will go to zero as the order of the polynomial (i.e., nth order polynomial is differentiable n times).This isn't perfect (see below), but to prove a function is infinitely (continuously) differentiable, say in some open interval, it is sufficient to prove that it has a convergent power series in that interval (the power series is then the Taylor series, and you can read off the derivatives). In particular, any polynomial works, as does the exponential, and sine and cosine.
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Indeed, the Taylor series of a polynomial is the polynomialUnless you consider taking the derivative of 0 as part of "infinitely differentiable" I don't think any polynomial works at all. In fact, no polynomial should work since the derivative will go to zero as the order of the polynomial (i.e., nth order polynomial is differentiable n times).
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Sure, the zero function is differentiable. Its derivative is also the zero function ((f(x+h)-f(x))/h = 0 for any nonzero h, so the limit as h->0 is zero). So it would have been a simpler example of a infinitely differentiable function, though perhaps not as interesting as the exponential :wink:
Infinitely differentiable just means that you can take as many derivatives as you want and they all exist (we're probably also requiring continuous derivatives of all orders).
Right
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Question: Are there any functions that are only finitely differentiable?
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Sure. Let f(x) = x^2 for x >= 0 and f(x) = -x^2 for x < 0. The derivative is twice the absolute value function, which is not differentiable at zero. So f is only once differentiable, not twice.
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Yeah but something tells me (VTBoy can probably clarify for certain) that they were asking for non-zero derivatives, which is a much more binding restriction than just having the derivative exist.
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I don't get this.Sure. Let f(x) = x^2 for x >= 0 and f(x) = -x^2 for x < 0. The derivative is twice the absolute value function, which is not differentiable at zero. So f is only once differentiable, not twice.
f(x) = x²
f'(x) = 2x
f''(x) = 2
f'''(x) = 0
f''''(x) = 0
f'''''(x) = 0
ad infinitum
:-k
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Perhaps. But "infinitely differentiable" is a well-used term that doesn't include that condition.
Eroica: the first derivative is the function f'(x) = 2x for x >= 0 and -2x for x < 0. So the second derivative is f''(x) = 2 for x > 0 , -2 for x < 0, and is undefined at zero, hence no higher derivatives can be defined at zero either.
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Yes we want a non zero derivative
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Hehe. Ok, forget nearly everything I've said :P
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