#help me with calculus

98 messages · Page 1 of 1 (latest)

placid panther
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I've been trying to find a counterexample for an hour for the following statement :
Given a function that is continuous in [a,b], and differentiable in (a,b). C is a point in (a,b). Then there are 2 points x1,x2 between a and b
a <= x1 < x2 < <= b so that
f'(c) = (f(x2) - f(x1)) / (x2 - x1)

abstract canopyBOT
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acoustic light
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For starters use the limit definition of f', then calculate what is (f(x+h)-f(x-h))/h when h tends to zero

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Now let f=sqrt(1-x^2) and find c such that f'(c) = 1. Now use this weird limit (fix it a bit) and observe what is happening to gain more intuition

dawn arrow
placid panther
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yeah but it's false

dawn arrow
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...but the mean value theorem is true.

placid panther
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i know but it's given that this statement is wrong in the question, and we have to give a counterexample

dawn arrow
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It's not, though, unless you wrote it wrong.

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But as written, it's just the mean value theorem.

placid panther
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i guess it should be for every c between (a,b) there are 2 points

dawn arrow
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...can you just take a picture or screenshot of the original text?

placid panther
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not really because it's in a different language, i tried my best to translate it

dawn arrow
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I expect the part you might have gotten wrong is in the mathematical notation.

placid panther
dawn arrow
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Before that.

placid panther
dawn arrow
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Where is c?

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Is it anywhere on [a, b]?

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Or on (a, b)?

placid panther
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no, it's anywhere on (a,b)

dawn arrow
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Then it can be between x_1 and x_2, which then just makes it the normal mean value theorem.

placid panther
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i guess i'll have to ask during the next lecture, thanks anyways

dawn arrow
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Okay, hang on.

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I think I might see the problem.

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Or wait, no. The counterexample I thought of is not differentiable on the required interval.

placid panther
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yeah..

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i don't even think the statement is wrong personally

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since we can always find such x1,x2 if we go closer every time to c

dawn arrow
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Or wait, I see how it's different from the mean value theorem.

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I don't know how that difference makes it false.

placid panther
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i asked chatgpt and it told me the statement is wrong but couldn't understand its counterexample

dawn arrow
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Literally all it knows is what sentences look like.

placid panther
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it really helps me though, it gives me the way to do stuff sometimes and i continue doing them, when im at a deadend

dawn arrow
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If you ask it something it can recognize as math, it'll ask WolframAlpha, but then that's still worse than you just asking WolframAlpha.

woven tapir
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put f(x) = exp(x) in [-1,1] and let c = 0. Then f'(0) = 1. On the other hand we must have (e^b - e^a) / (b-a) = 1 for some a < b in (-1,1). This can't happen. For otherwise e^b-e^a = b-a

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@placid panther

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for intuition, if you allow rolle's theorem to hold, then you won't find a counterexample

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so try something strictly monotone

placid panther
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alright, thanks! i'll try it

woven tapir
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as for why said equality can't hold: the only way this holds is if b=a=0

acoustic light
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I mean the theorem is just another way of saying that a secant line parallel to a tangent cuts the function on at least two points in the interval

woven tapir
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apparently it isn't

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the statement allows one to pick c freely

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that's the problem with it

acoustic light
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And e^b-e^a = (b-a) has solutions with the lambert function???

woven tapir
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the only real solutions are 0

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you can probably conjure something with analytic continuation, but that's wandering a bit too far above our paygrade here

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unless i made a mistake, you're free to correct

acoustic light
woven tapir
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that line of thinking wont work here

acoustic light
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A numerical example that of course is not exact but you can try solving for

e^x = x+1.005, this is the secant line 0.005 above the tangent line for x=0, f(x) = e^x

woven tapir
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you can prove directly that only a=b=0 works

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being able to get (arbitrarily) close to 0 is not sufficient here

acoustic light
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Okay so you're saying that if you give me, say, a = -1/2, I can't find b?

placid panther
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really? that was my problem with the question (other than finding a counterexample i mean), i thought that there are always x2 and x1 that can work for every c, since the MVT would work once the x2 and x1 are really close to c

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or maybe you guys are talking about different stuff i got no idea my brain stopped when big words came in chat

woven tapir
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the mvt states that there exists some c that fits

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says nothing about arbitrarily chosen c

acoustic light
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For a = -1/2 I found an explicit solution for b.

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So you are not correct that only a=b=0 works

woven tapir
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hm, indeed, I made a logic error

acoustic light
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So I guess my point stands:

Pick a tangent of f(x) at point c, this is possible because f was differentiable. Now adjust the tangent to become a secant by epsilon. The points of intersection of the secant and f(x) are precisely x1, x2.

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Or in other words, if you give any x1, x2, then I can find epsilon.

placid panther
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hmm

woven tapir
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you pick c and you have to find x1 x2

acoustic light
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There are infinitely many different x1, x2

woven tapir
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thought about it some more, If we assume continuously differentiable then I believe what you say works

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your argument basically says there's a secant line of large slope, and another secant line of small slope, then you squeeze around c and appeal to continuity

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but continuity of the derivative

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not f

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@acoustic light@placid panther

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at the very least, continuously differentiable around c

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so we must look for counterexamples among non smooth functions

placid panther
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thanks guys, ill think about it tomorrow when i wake up

acoustic light
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And if f is once differentiable on [a,b], then f' is cont...

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So f doesn't need to be smooth

woven tapir
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f(x) = x^2 sin (1/x^2), f(0) = 0, is a classic example

woeful anchor
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What about just, say, sin(x) between -pi/4 and pi/4 and c = 0?

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since I think the slopes (sin(b) - sin(a))/(b-a) would be strictly upper-bounded by sin'(0) = 1

woven tapir
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without mvt:

$$ |\sin b - \sin a| = 2\left\lvert \cos \frac{b+a}{2}\sin\frac{b-a}{2}\right\rvert $$

night roseBOT
woven tapir
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but then it's leq 1, is it strict actually?

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oh wait I think sinx < x is strict for nonzero x

woven tapir
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@acoustic light @placid panther

woeful anchor
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it is the case if you consider convexity arguments i think

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whether or not a and b are both of the same sign

woven tapir
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$$ \left\lvert\frac{\sin b -\sin a}{b-a}\right\rvert = \frac{1}{b-a} \left\lvert\int _a^b \cos t dt \right\rvert \leqslant \int _0^1 |\cos (a+s(b-a))|ds < 1 $$

night roseBOT
woven tapir
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k now I'm convinced

placid panther
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Ggs, thank you all

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+close