This is a dumb question. In school you learn that f''(x)

>>7682782
no, thats exactly what it is. the curvature.
whats the problem here?

Just because it LOOKs like it's 0 in a wide range doesn't mean it actually is. That's just the limitations of visuals. x^4 is only 0 when x = 0.

Anyway "curvature" is a rather nebulous description. The derivative of a functions more often explained as the slope of the curve at a certain point (in which case the points near 0 would be CLOSE to 0, but not exactly 0 except at 0). Or, more generally, the change in the value of f(x) with respect to the change in x (which is actually its definition).

>>7682786
My problem is that I don't see how x^4 differs from x^2 in terms of curvature.
Both tend upwards both ways from x=0.
So how can the curvature of x^4 be 0 (when that of x^2 isn't)?

>>7682790
I know, this is what I mean. The curvature of x^4 only looks as thought it was 0 at x=0, but in reality it isn't.
But f''(0)=0
Whereas for f(x)=x^2, f''(x)=2.
What's the fundamental difference between x^4 and x^2 that explains this?

>>7682794
f'(x) = 2x

>>7682800
I'm asking about f''(x).

>>7682800
>>7682794
Oh, there's another tick there. Damn font.

Anyway, the double derivative is the rate of change of the rate of change. It's not exactly something that's easy to visualize. There's no particular contradiction and I'm not sure what the issue is.

The double derivative of x^4 is 0 at 0 because at 0 its growth rate is not growing. The double derivative of x^2 is 2 at 0 because at 0 (and, in fact, at all points) its growth rate is growing.

>>7682794
>What's the fundamental difference between x^4 and x^2 that explains this?
The degree of the polynomial.

>>7682809
Why is the growth rate of x^4 not growing at 0 when that of x^2 isn't? This is what I don't understand.

The curvature is actually [math] \frac{f''(x)}{(1 + (f'(x))^2)^{ \frac{3}{2} }} [/math]

>>7682813
Because x^4's double derivative is 0 at 0 and x^2's isn't. That's the reason that we have derivatives in there first place: because visuals can be deceptive, but math doesn't lie.

They're different curves. They might have certain similarities, but the complication of their growth rates are different.

>>7682813
It just is like that. I mean, they are different functions so they behave differently.
I don't feel like there is something deeper here.

>>7682791
They both go down, flatten out, then back up again, but not in the exact same fashion.
The gradient of x^2 IS zero at x=0, so I'm not really sure what you're saying.

>>7682812
>>7682816

What I'm really trying to ask is:
Both f(x)=x^2 and f(x)=x^4 describe graphs that tend upward in both directions at increasing rates. The growth rate of x^4 is higher for larger x, just as it is for x^2.
If f''(x) describes the growth rate of the growth rate, what property of those functions justifies that being >0 for x^2 and 0 for x^4?
I know it's the degree of the polynomial, I'm also aware what the first derivatives look like.
What I don't see is how the growth rate of the growth rate of x^4 can actually be 0 in that point.

>>7682825
I'm asking about the second derivative, which supposedly describes how fast the function changes its growth rate. The result of 2 for x^2 implies that its growth rate is changing.
The result of 0 for x^4 implies its growth rate is not changing. But it is, because it grows faster the further away you get form x=0. It seems paradoxical to me.

>>7682828
>If f''(x) describes the growth rate of the growth rate, what property of those functions justifies that being >0 for x^2 and 0 for x^4?
Because x^2's growth rate is constantly growing whereas x^4's growth rate is growing at a rate of 12x^2. There is really no better or more intuitive way to describe the way a function's growth rate behaves than its derivative.

>What I don't see is how the growth rate of the growth rate of x^4 can actually be 0 in that point.
The important thing to remember about the point x = 0 is that it is only a point. Everywhere around x=0, the growth rate is growing positively (since x^2 is positive at everywhere but 0), but at that exact point the growth rate stops (but just for a point). A derivative being zero at a certain point can mean a lot of things such as local max or a change in curve orientation. In this case 12x^2 being 0 at 0 means that 4x^3 changes from curving down to curving up at 0 and 4x^3 being 0 at 0 means x^4 has a local minimum (and, in fact, an absolute minimum) at 0.

>>7682841
I accept that. But then the prose explanation you get in school that f''(x) = curvature, or how concave the graph is in that point seems lacking.
Geometrically, in a point where the function increases on both sides, it logically has to be left curved, no?

>>7682851
I would interpret it that way: The growth "slows down" around 0, which leads to the big "belly" of x^4. It's not very elegant, though, but it should work as a simple explanation.

>>7682851
> f''(x) = curvature, or how concave the graph is
> in a point where the function increases on both sides, it logically has to be left curved
the "concaveness" of x^4 changes around 0, at 0 it isn't concave anymore, but then it starts to get more and more concave, that's why the minimum of x^4 looks so flat.

>>7682782
>f''(x) describes the curvature
No, it does not.
f"(x) is the rate-of-change of slope, which has nothing to do with curvature.
Curvature is useful, therefore Lrn2curvature.

>>7682918
Sure it does.
See the formula here:

>>7682816

>>7682782
The shape of x^4 around x=0 is quantitatively flatter than that of x^2.
How you quantify this difference is exactly in taking the second derivative and plugging in x=0.

>>7682918
Semantincs.
Of course, the curvature of f(x)=6 is 0 for all x, so there is no "curvature". This word is a name for a concept that's made into a function here. We can call it zugbux if that helps.

I guess I understand it and the concept of "curvature" is just sort of muddy here. I still wish there was some deeper explanation of why x^2 behaves one way (nonzero growth of slope at 0) and an infinite number of higher even-order functions x^4, x^6 etc do not, when they all essencially do the same thing (multiply x by itself n times).

>>7683046
because x^4 is flatter than x^2 at the bottom, this means that x^4 is less and less curved as it approaches 0. x^2 on the other hand hasa constant curvature for all x values.

>>7683064
But isn't it only "flatter" because x <1 get multiplied with themselves more often, thus producing smaller values? Basically just the inverse of its faster growth for x>1. X^2 also makes x>1 smaller, only less so. Just the same as x^4 is flatter than x^6 without any implications like this. I see a quantitative difference, but it still doesn't seem intuitive that a basic property is different for x^2 from all other even degree functions.

>>7683080
>But isn't it only "flatter" because x <1 get multiplied with themselves more often
This, in consequence, leads to a curvature that gets smaller as you approach 0.
I don#t know what the problem is here, x^2 having a positive curvature at 0 seems to be intuitive to you, but in reality, it is the exception, there are infinite funktions x^n with even n values, that have a curvature of 0 at 0.

>>7683090
I'm asking why the exception exists. I understand the idea of the function becoming flatter. What's unintuitive to me is that x^2 differs x^4,x^6 etc. when it's fundamentally the same thing.

>But isn't it only "flatter" because x <1 get multiplied with themselves more often
This, in consequence, leads to a curvature that gets smaller as you approach 0.

This should apply equally to x^2.

>>7683121
no, as you multiply x only once with itself, so you uniformly curve the line. When you now multiply x again and again to itself, the curvature will get smaller and smaller (for 0<x<1).

>>7682782
Imagine running on the x^2 curve from negative to positive Xs. You'd be always be tilting towards your left side while running, even at x=0.
While you instead run on the x^4 curve, for a brief moment at x=0, you be running on a straight path.

>>7683141
yes, that's what you do as the curvature gets closer and closer to 0. Is that so hard to imagine? You're just going on a straight path in an infinitesimal small interval, so you won't notice it.

>>7683141
>>7683145

No, I really don't get this part.
Why am I tilting left when walking on x^2 but not walking on x^4? Standing on 0, the infinitesimally closest x is greater than 0. I'm thus stepping left. Standing infinitesimally close to 0 on the negative, x is also greater than 0, thus I am also stepping left. How is this different between x^4 and x^2?

>>7683166
the picture of going in a straight line for one point is so flawed. you're tilting left, but at each step you tilt a little less left until you almost go straight, then you reach 0 and start to go more and more left again with each step. At 0 itself you don't go in a straight line, since it is only a single, infinitely small, point. It's impossible to go straight at a point, since a point isn't a distance, but the boundary value of the curvature is 0.

>>7683176
I'm not bothered by infinitesimal steps. I understand that you're never "walking in a straight line" when passing an inflection point either.
What seems unintuitive is that this should be the case when evidently, all
infinitesimally close points lie to the left.

But aside all that: What sets x^2 apart from x^4,6,8 etc?

On x^2, if I'm standing on 0, all points on front of me lie to the left. Same for x^4. What makes x^2 special?

>>7683166
The centripetal force you need to keep walking on the x^4 curve would be exactly zero at x=0

File: 1446240080808.jpg (15KB, 223x233px) Image search: [Google]

15KB, 223x233px

>>7682832
>second derivative
>result of 2 for x^2
>result of 0 for x^4

>second derivative of x^4
>zero

>>7683274
Evaluated at x=0 you flying fuckwit, like holy shit is it that painstakingly hard to understand? Goddamnit.

>>7682782
replace y with f(x) and plot some numbers.

>>7683368
pls read the thread

>>7683256
Why not for x^2?

>>7683241
I really don't know your problem, anon. Please see a doctor to have your autism checked.

>>7682832
They are different though

>>7683425
because it's more curved.

>>7683425
you sound upset

>>7683452
should link to >>7683390

>>7683452
x^4 is also more curved than x^256, what makes x^2 unique?

>>7683460
>what makes x^2 unique?
That it's x^2? What makes 2 the only even prime?

>>7683462
the fact that all other even numbers can be divided by 2 without being divided by themselves.

I'm wondering why this question seems so weird to many people. You have an infinite number of functions that multiply x by itself a number of times. They all have 0 curvature at 0. Then you have one function that does the same thing and whose graph moves down, then up just like all the others. But it has a curvature of 2. I find this odd.

>>7683460
x^2 is the only exponential function that is so curved, that it's curvature at 0 is bigger than 0.
All other exponential functions are less curved at x=0 because of numbers and math and shit.
At this point I don't even think you're the same guy who started it, there are probably 4 or 5 different people asking "but why not x^2?" again, each time someone replies, just to troll people.

>>7683471
No, I'm the same guy, I'm still kinda waiting for a satisfactory answer on why "x^2 is the only exponential function that is so curved, that it's curvature at 0 is bigger than 0."
"because of numbers and math and shit." doesn't cut the mustard. It's not a troll question and I'm not trying to waste anybody's time. Just checking the thread in case someone has a good explanation.

>>7683475
>x^2 is the only exponential function
It's a polynomial you utter fucktard

>>7683475
it is the only polynomial that has a constant curvature, since its x is only multiplied with itself one time, while all others' x's are multiplated with themselves multiple times, thus changing the curvature for each x and making it less curved around 0.

>>7683477
You may have noticed those quotation marks around the sentence. In english, quotation marks are used to quote something someone else said (this is, actually, why they are called such). I don't think you are very intelligent.
(You're right though, I overread the mistake).

>>7683482
How new are you? We use
>memearrows
here

>>7683477
Look at that, you're right. I will try to remember that.

>>7683487
I'm >>7683471 btw, and started calling it that way.

https://www.math.hmc.edu/calculus/tutorials/secondderiv/
Here you go

>>7683481
I guess this makes *some* sense to me, but not much. I understand why each higher degree makes it less curved around 0. But only at 0 does the curvature "suddenly" jump from nonzero to zero when you go from x^2 to x^4. I don't really see why multiplying by x an additional time has this effect. The infinitesimal vicinity of 0 is "infinitesimal*infinitesimal" for x^2, which constitutes a positive curvature. But "infinitesimal*infinitesimal*infinitesimal" doesn't. Maybe it's just me being weird but I don't really see the correlation.

>>7683490
if you actually believed that page contained a proper answer to the question at hand, you'd have pasted it.

>>7683493
before anyone gets anal about it
>But "infinitesimal*infinitesimal*infinitesimal" doesn't.
should be "infinitesimal*infinitesimal*infinitesimal*infinitesimal" since x^4

>>7683493
well, if you just "jump" from 2 to 4 in the exponent, the curvature can "jump" from 2 to 0. The problem is, that there are no polynomials for exponents other than natural numbers, so the curvature doens't go smoothly from 2 to 0 when you change the exponent smoothly from 2 to 3 (since the second derivative of x^3 at x=0 would already be 0).

>>7682782
>This is a dumb question.

yes it is. dont worry, it will pass.
dont worry, be happy ^_^
math is hard, but mathgasm is worth the effort.

>>7683499
It's not so much the jump itself that irritates me, but rather the fact that I don't see the feature of x^2 that distinguishes it from all the other functions. Each higher degree function is "less curved" than its predecessor, but they don't change in any basic properties like this one. Why does "multiplaying x" any even number of times down to 4 do the same thing, but multiplying it only once removes curvature?

This seems to me just as strange as supposing that dividing a function by any arbitrarily large factor at some point makes it have 0 slope

>>7683501
fortunately, it's not anywhere near as dumb as your answer

>>7683493
So it seems like you understand why x^n for general natural n has zero curvature at x=0.
So let's just focus on why x^2 doesn't.

Looking at x^0, we see a flat line. Think about f(x) as position, x as time, permitting negative times.
Then x^0 indicates that we aren't moving. At all. Ever. So, no curvature.
For x^1, we find a slope. So as we go forward in time, we move in position. Now we have a velocity.
For x^2, we have an acceleration. Now, this is a uniform acceleration that does not depend on x at all. Its a car accelerating at 2 whateverwhatever/whenevers constantly.
Now, as soon as we go up in degree by one (or two) its not a constant acceleration. The car accelerates faster at larger x.

The x^2 case doesn't have zero curvature at x=0 precisely because its curvature gives no fucks about its position.

>>7683506
As for the arbitrary number thing, this one is easy to demonstrate, but let's be careful with our words.

Imagine some function. Then this function somewhere has a non-infinite slope (forgive me). Around there, pick an interval and look at its rise over run (kind of like slope).
Now divide its rise by a huge-ass number. BOOM! Its kind-of-slope just dropped drastically.
Using this, you can get as close to zero as you'd like. (For every epsilon>0 around zero, you can pick a big enough natural N...)
See?
So a lot of people just like to say that in the limit the slope goes to zero.

You probably mean the right thing, but the derivative is the gradient (or slope) of the function, not the curvature.

>>7683510
this right here
this is the answer the OP asked for

>>7683510
Indeed this is a really good explanation. Even though I knew this already, I think my definition of curvature was just lacking. You can't always equate it with the "bend" of the actual graph I guess. Looking at it strictly as the change rate of the previous derivative makes it much more flexible.
>>7683521
Using a wider definition of curvature than "is it a left or right curve" I guess it's also more feasible to interpret the 0 as a limit.
Thanks, this makes sense, props.

>>7682782

seems like the problem may arise from OP's arbitrary choice of derivative.
The second derivative of x^2 is 0, and similarly, the nth derivative of x^n is zero.

>>7683688
f: x-> x2
f': x -> 2x
f": x -> 2
f"': x->0
so the (n+1)th derivative of a polynom nth order is 0

>>7682828
Between 0 and 1, the higher the power is then the smaller the result is. e.g. 0.5^2 = 0.25 and 0.5^4 = 0.0625 which appears closer to 0 and gives x^4 a flat look at the bottom

derivative =/= curvature
moving along the cure at a constant speed =/= moving along the x-axis at a constant speed

this is the source of your confusion

>>7682794
change the window on your graph and you'll see that it's only 0 at x = 0

>>7683005
>Semantincs
>distinguishes you from the other apes
don't misunderestimate it

>>7682832
Okay I see what you're fucking up in your head op. The rate of change of the rate of change of the rate of change of the growth rate is 4! So the rate of change of the growth rate is changing

Try not to be so autistic op.

>>7683688
>The second derivative of x^2 is 0

no it's not

>>7682782
f''=12x^2

put any x here and get the curvature. works for both even and odd polynomials dude

>>7682813
x^2 is always shaped like a U. x^4 isn't shaped neither U nor n at x=0