4 ms·
Went from starting a car analogy to this, as part of the simple car analogy: > Now the mean square of something that deviates around an average, as you know, i
by Supermancho 7y ago
Went from starting a car analogy to this, as part of the simple car analogy:
> Now the mean square of something that deviates around an average, as you know, is always greater than the square of the mean; so the kinetic energy integral...
No, I don't know, and you lost me. I don't think it matters to me anyway.
- siddboots 7y agoI love the Feynman lectures, but I feel like he does this sort of thing all of the time. He's so keen on delivering intuition for the subject that sometimes he glosses over crucial technical details.
- username90 7y agoThe problem is that to get intuition for A, you also need to have intuition for all of A's dependencies. In this case if you had good intuition for high school math the statement would be obvious, but almost nobody build that kind of intuition.
- absherwin 7y agoOne bit of additional intuition: Since the square of a number goes up from addition by slightly more than it goes down from subtraction, perturbations increase the average of the squares. This is why the difference between the two quantities Feynman mentions is used to measure the variance of a set of numbers. Since that’s still not precise, let’s compare the square of the mean to the mean square for two numbers a and b. The square of the mean is ((a+b)/2)^2=(a^2+2ab+b^2)/4 The mean of the squares is (a^2+b^2)/2 Feynman’s claim is that the second is always bigger if the numbers deviate around an average (a and b aren’t equal). So let’s subtract the first from the second. We get (a^2-2ab+b^2)/4. The numerator is equivalent to (a-b)^2. Since a square of a real can’t be negative, when a and b are unequal the mean of the squares is always larger.
- mobileexpert 7y agoSee also https://en.m.wikipedia.org/wiki/Jensen%27s_inequality https://en.m.wikipedia.org/wiki/Jensen%27s_inequality
- hardlianotion 7y ago1/T \int 0_T x(t)^2dt >= (1/T \int 0_T x(t) dt)^2
- heisenbit 7y agoYou likely know that driving constant 60 is using less gas than driving 30 half and making it up with 90. While is not a square function it may be close enough for a rough real life analog to help.
- happycube 7y agoNot just that, but there's (slightly) less effort involved in keeping the car at 60.
- Supermancho 7y agoAh, I get that. Thank you.
- fgonzag 7y agoaverage of squares > square of averages
- Yajirobe 7y agoIt is also wrong. If variance is zero, then the mean of the square EQUALS the mean squared.
- dagw 7y agoYou could argue that if something _deviates_ around an average then its variance cannot be zero. But I agree that the language could be more precise.
- evanb 7y agoSuppose you have a list of N numbers x_i, whose mean is X. The unaveraged variance, ∑_i (x_i - X)^2 is the sum of a bunch of squares, which are non-negative things, so the variance is non-negative. Now, we can expand the square 0 ≤ ∑_i (x_i - X)^2 = ∑_i x_i^2 - 2 X ∑_i x_i + X^2 ∑_i 1 ≤ N * (mean of the squares) - 2 N X^2 + X^2 N ≤ N * (mean of the squares) - N * (square of the mean) and find square of the mean ≤ mean of the squares equality being saturated if and only if the list is a constant list.
- jpeterson 7y agoI'm guessing he's relying on common knowledge of variance here. If you know that variance is always nonnegative, and you also know that variance can be defined as (mean square - square of the mean), then it's obvious.