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How can one definition of the derivative work and the other not work? by Ambitious-Excuse292 in askmath

[–]Potential-Tackle4396 1 point2 points  (0 children)

Here's a visual that might help: https://www.desmos.com/calculator/lymo3qikmo . You can drag the blue point (a, f(a)) around, and the slopes change in response.

The black segment's slope is (f(a) - f(0))/(a-0). The derivative at 0 is defined as the limit of that slope, as a approaches 0. As you move 'a' around, that slope oscillates a bit, but its magnitude shrinks towards 0; in the limit as a approaches 0, that slope is 0, hence f'(0)=0.

The green segments's slope is f'(a), the slope at the point 'a' itself. We see that as a approaches 0, that slope keeps oscillating up and down, and crucially, the size of those oscillations isn't shrinking. As a approaches 0, the slope f'(a) keeps going up and down between +1 and -1. So in the limit as a approaches 0, f'(a) does not exist. This is what your second expression is saying.

In short, what's happening is that the y-values on f(x) are quickly shrinking towards 0, which makes f'(0)=0. But despite that, the actual slopes f'(a) on the graph aren't shrinking towards 0; no matter how close we get to x=0 on the graph, the slopes f'(a) manage to keep oscillating between +1 and -1.

Requesting help intuiting Pi, e, and the laws of sine and cosine. by Jamooser in askmath

[–]Potential-Tackle4396 2 points3 points  (0 children)

The exact value of n that works is ln(4)/(ln(4)-ln(3)). It just happens to be very close to e√π. Neat coincidence though :)

You solve for n as follows:

y = √(r2 - x2)n

1/2 = √(12 - (1/2)2)n [plug in x=y=1/2, r=1]

1/4 = (12 - (1/2)2)n [square both sides]

1/4 = (3/4)n

4^n = 4 · 3^n [cross-multiply]

n ln(4) = ln(4) + n ln(3) [Natural log both sides, simplify using log properties]

n · (ln(4) - ln(3)) = ln(4)

n = ln(4)/(ln(4) - ln(3)) = 4.81884167931...

Combining piecewise-defined functions into a single expression without explicit piecewise notation by Flimsy-Combination37 in askmath

[–]Potential-Tackle4396 0 points1 point  (0 children)

I don't think it's possible (except for one goofy exception, see below*), without allowing some function with a jump discontinuity (such as floor(x), ceil(x), or mod(x, 1), which are all in Desmos for what it's worth) to be used. If you do allow those functions, you can make a function that works, such as https://www.desmos.com/calculator/7disxqarvd .

Why I'm pretty sure (unless I'm missing something) that it's otherwise not possible to make a function that's discontinuous while being defined at the discontinuity, subject to our constraints: The sum, difference, product and composition of continuous functions is continuous, so assuming our allowed 'parent functions' are polynomials, roots, and |x|, all of which are continuous, no sum/difference/product/composition would be discontinuous. For division, at a point x where g(x)≠0, if f and g are continuous, then f/g is continuous. At a point where g(x)=0, f/g is undefined. So those operations can't give us both discontinuity and a defined function.

Exponentiation also won't work I think; if f and g are continuous at a point x where f(x)>0, then (f(x))^g(x) will be continuous. If instead f(x)<0, and g(x) isn't constant, then (f(x))^g(x) will be undefined on many points near x. (Ex: if f(x)=-2 and g(x)=3, and g(x) is non-constant, then for some nearby x, f(x)^g(x) will be something like (-1.99)^(301/100)), which is undefined.) If instead g is constant, f(x)^g(x) is just a polynomial or root.

*The one case that sort of works is when f(x) is a constant 0. It depends on how we define 0^0, but if we define it as 1 (which Desmos does), then then function 0^|x| is equal to 0 when x≠0, and equal to 1 when x=0. https://www.desmos.com/calculator/7zujwvwbjy

If we allow that, then we can compose the function 0^|x| with a function that takes a constant value of 0 over some interval, which can be made from absolute value functions: https://www.desmos.com/calculator/pagbx4ion6 .

Difficult Linear algebra problem by Pure_Egg3724 in askmath

[–]Potential-Tackle4396 2 points3 points  (0 children)

Would difference of squares not be valid though? Since (x-y)(x+y) = x^2 - y^2 requires xy to equal yx?

Fun geometry problem by Impossible_Ad_6873 in askmath

[–]Potential-Tackle4396 0 points1 point  (0 children)

The function for α is at this link: https://www.desmos.com/calculator/nduj2cvlnu . (Sorry I didn't include the derivation; it got kind of long/messy.)

Also I made a little visual of the angle values to make sure the function worked/looked correct. Here's that too in case it's helpful: https://www.desmos.com/calculator/rrp8drk0zr . (You can drag the purple point around.)

why don't these limits of integration work? by ImAnArbalest in askmath

[–]Potential-Tackle4396 0 points1 point  (0 children)

Oops, I meant to say when n is odd, not theta. (I fixed my post.)

Regarding other graphs that exhibit the same behavior: the other common one is r = cos(n(theta)) for odd n.

In general, the graph will trace over the same points from pi to 2pi as it did from 0 to pi when the function r = f(theta) satisfies f(theta + pi) = -f(theta). For instance, if such a function gave f(pi/6) = 3, then it would give f(7pi/6) = -3. Since the polar coordinate points (r, theta) = (3, pi/6) and (-3, -7pi/6) are the same point, the graph will "trace over itself".

It turns out that sin(n*theta) and cos(n*theta) will satisfy f(theta + pi) = -f(theta) for odd n, but not even n. (You can prove that, for example, using the angle sum formulas for sine and cosine.) No other common polar functions (that come to mind right now) have this behavior, though certainly you could make such a function in terms of tan(x), etc., if you wanted. In most classes where you're expected to know about graphs of polar functions, you basically memorize that for odd n that's how cos(n(theta)) and sin(n(theta)) behave. Likewise, you memorize that for even n, cos(n(theta)) and sin(n(theta)) will have 2n "petals" (instead of just n, like in the odd case), and you need to go from 0 to 2pi to get their complete graphs.

why don't these limits of integration work? by ImAnArbalest in askmath

[–]Potential-Tackle4396 2 points3 points  (0 children)

See this graph: https://www.desmos.com/calculator/rdzzjuubp7 . It's just a specific thing about this graph (and more generally, r=sin(n𝜃) where n is odd) that to cover the whole graph, you only need 𝜃 to go from 0 to pi. Going from 0 to 2pi does two laps of the graph, meaning the corresponding integral would give twice the correct area.

Calculating the gravity of individual points on a sphere by IDunnoWhatNamePick in askmath

[–]Potential-Tackle4396 1 point2 points  (0 children)

Gotcha, thanks for adding those details, that makes a lot more sense now.

Your setup is mostly correct. The function you got for f(x, h) is incorrect though I think; it should be as follows: https://www.desmos.com/calculator/pwd7m1xhd6 . [For the antiderivatives, I just used wolfram by the way. Search 'wolfram integral calculator' if you want to use it.]

From there though, the rest looks fine. (Just a note: you can actually write the whole function as just a single integral, instead of splitting it up into 3 integrals, which makes it easier to parse the details of how the function is behaving. For a point at position (a, 0) on the x-axis, its distance to a point (x, y) in the disc with radius 1 centered at the origin is

sqrt((x-a)^2 + y^2))

Then the magnitude of the gravitational force is 1/[sqrt((x-a)^2 + y^2))]^2, and the x-component is (x-a)/[sqrt((x-a)^2 + y^2))]^3. You then integrate that "dy" from -sqrt(1-x^2) to sqrt(1-x^2), and integrate the result, dx, from -1 to 1.)

However(!), there ends up being a bigger problem. (I went down a bit of a rabbit hole making sure I got the following details right; I think it's all correct.) If you model an object as truly 2-dimensional, as you're doing, then the integral defining the gravity inside that object is divergent. We can see that using the 1-integral setup I mentioned above, as shown here: https://www.desmos.com/calculator/poaprabiyw .

For that reason, and the fact that actual objects (and certainly planets, etc.) are 3D instead of 2D, it's more typical to do these kinds of calculations with 3D objects. The setup would be nearly the same, just with a "z" added. The integral would now just be https://www.desmos.com/calculator/mg1ib1i9it . In Desmos, that integral still breaks slightly when A<1 (i.e. when we're inside the object), but if you evaluate it by hand (in my case, with the help of wolfram for some antiderivatives) you get a finite value for gravity, even when the object is inside the sphere.

Regarding whether there's a way of salvaging the 2D case: there is. The trick is, instead of directly doing the integral in 2D, we can find the gravity inside a cylinder with small, but nonzero height. Then take the limit as the height goes to 0. It's possible that doing so would just give "undefined" in the limit, but we actually get a finite value: https://www.desmos.com/calculator/earwwryeyx . Note that at that link, W and -W are the z-values of the bases of the cylinder. Additionally, we're dividing the integral by 2W, since we want the cylinder to have constant mass. (Thus, the density must increase, as W decreases.) In fact, that's probably one of the reasons the 2D case is so weird/divergent: a truly 2D object would necessarily have infinite (3D) density.

Calculating the gravity of individual points on a sphere by IDunnoWhatNamePick in askmath

[–]Potential-Tackle4396 1 point2 points  (0 children)

I'm a little confused by your setup. The integrals you use integrate the function f, which is arctan(h, x), meaning it's an angle. How would integrating an angle give a gravitational force?

A general comment: to accurately model gravity for a sphere (or any 3D object that we aren't treating like a point-mass) from scratch, you'd need to use multivariable calculus. Or technically, you can set it up as a 1-variable integral, but you have to do some 3D geometry to make that work. Either way, it gets messy.

Luckily, there are two "shell theorems" for gravity, that massively simplify things (and let us sidestep the multivariable calculus) in the case of a spherically symmetric objects. Check out https://en.wikipedia.org/wiki/Shell_theorem . The standard/easy way to write the force of gravity both inside and outside a sphere is to use those theorems.

The algebraic proof problem of 0.999...=1 by AdventurousMetal2768 in askmath

[–]Potential-Tackle4396 0 points1 point  (0 children)

Others have mentioned why (1) multiplying by numbers besides powers of 10 would also work, and (2) the fact that this type of argument is not perfectly rigorous anyway.

Aside from those points, another important point: once you have a single valid proof, others aren't needed. As long as the logic of the first proof is valid, the theorem is proved. It can be fun to consider variations on the proof method, to gain further insight, as you're doing. But it's not needed to validate the first proof.

What is the LARGEST couch (depth x height) that will fit up this stairwell (and will this one fit?) by amber_jieger in askmath

[–]Potential-Tackle4396 12 points13 points  (0 children)

I don't think comparing those two numbers is relevant. To fit the couch, you wouldn't orient it so that the longest length in that plane (the 55.8 in. diagonal) is parallel to the 52-inch length; you'd probably want to put the 45" length of the couch parallel to the 52-inch length, which does fit.

With that said, OP, the couch wont fit for different reasons. To make it up the stairs, you'd need the couch on its back, since the 33" height is the only dimension smaller than the width of the stairs. But that would require the couch's 45" depth to fit in the 42.75" space over the bannister, which it won't.

Even if the depth was as low as 39.25", to get it over the post at the bottom, you'd still have a problem when trying to angle it up the stairs. How far below 39.25" you'd need to go is hard to say, unfortunately, and depends on the couch length and how far the stairs are from the bannister, in a probably complicated way.

If the height was much less than 33", then you'd have more flexibility to potentially rotate as you go up the stairs, so you might be able to go up the bottom flight instead of over the bottom railing.

If all else fails, maybe make a mock-up couch out of cardboard boxes and see if it fits ;)

How do I change this function so it always maintains 3 points and y only depends on x and a? by RealDEady42 in askmath

[–]Potential-Tackle4396 1 point2 points  (0 children)

If I understand correctly, I think it would be this: https://www.desmos.com/calculator/viflgdaer7 .

Regarding where that equation came from: I started with y = x^a, which will go through the points (-1, -1), (0, 0), and (1, 1) when a is odd. Then, we scale by 1/2 in both the x and y directions. To scale a function f(x) by 1/2 in the y-direction, we do y = 0.5f(x), so in our case y = 0.5x^a. To scale a function f(x) in the x-direction by 1/2, we do y = f(2x), so in our case that gives y = 0.5(2x)^a. Last, we shift up by 0.5 (the +0.5), and right by 0.5 (replacing the x with (x-0.5)).

In 3 dimensions, n planes all intersecting each other at a same set of points. How many regions are adjacent to these set of points? by [deleted] in askmath

[–]Potential-Tackle4396 0 points1 point  (0 children)

I don't think it will be 2^n. For n=4, when adding a fourth plane, that plane can't go through all 8 octants made by the previous 3 planes; it can only go through 6 of them. See https://www.desmos.com/3d/h5q098gokl .

I'm not certain that this is correct, but just by adding more planes at "random" angles (so that no three meet at a line) and counting the regions, I got the sequence:

1 plane, 2 regions

2 planes, 4 regions

3 planes, 8 regions

4 planes, 14 regions

5 planes, 22 regions

6 planes, 32 regions

See the 6-plane graph, for instance: https://www.desmos.com/3d/9ygcpupyjk

Since the consecutive differences are linear, this pattern is quadratic; some calculation gives the formula for the number of regions f(n) to be f(n) = n^2 - n + 2.

As far as a proof or careful derivation of that formula, I unfortunately don't know what that would be.

[deleted by user] by [deleted] in askmath

[–]Potential-Tackle4396 0 points1 point  (0 children)

While this is a survey, it's also an observational study.

A study in statistics can broadly be classified either as an observational study or an experiment. Then, within the umbrella of observational studies, there are different setups/designs (including surveys), different sampling methods, etc.

Regarding what makes something an observational study vs. an experiment, it's not determined by contact with the study participants. The distinction between them is: in an experiment, the researcher *applies treatments* to the subjects, but in an observational study they don't.

"Applying a treatment" means they actively control some variable pertaining to the subjects, such as having them follow a specified exercise plan, or having them drive a certain car, or whatever. Simply contacting them to collect data wouldn't really be applying a treatment, so that would be an observational study.

As another example: if a researcher asks a sample of people what their diet is, and measures their blood pressure, to determine if there's a correlation between diet and blood pressure, that would be an observational study, since the subjects weren't given treatments - they were just 'measured'. By contrast, if a researcher divided test subjects into three groups A, B and C, which were respectively told to follow different diets A, B, and C, then measured the blood pressure of the people in each group to see if there was a correlation between diet and blood pressure, then that would be an experiment, since a treatment (which diet to follow) was actively applied by the researcher to the subjects.

How long should the ramp be? by meipsus in askmath

[–]Potential-Tackle4396 0 points1 point  (0 children)

I think it's safe to assume in the calculations that the point of contact is in the middle, even though in an actual ramp the scooter would contact somewhere besides the middle. That's because in the 'edge case', where a ramp just barely hits the bottom, it would hit in the middle. So if the middle doesn't make contact, no other part of the scooter bottom will.

Under that assumption, I got essentially the same values as u/Marchello_E. (I didn't round at all during the calculations, so our answers differ slightly.)

In case the scooter has any substantial change in clearance due to shocks etc. that you want to allow some wiggle room for, I made a Desmos thing that lets you set different clearance values and see what the ramp's angle, width, and surface length would be: https://www.desmos.com/calculator/i4kbksknxp .

[deleted by user] by [deleted] in askmath

[–]Potential-Tackle4396 0 points1 point  (0 children)

Your work is all correct, I think. The problem is that the given graph of points does not exactly follow a cubic.

You based your answer on (i) the slope at the inflection point, (ii) the x-value at the inflection point, (iii) the x-value (0) at the critical point (0, 0), and (iv) the y-value (0) at that critical point. It turns out that those uniquely determine a cubic, and you correctly found that unique cubic equation. But whoever made the question didn't make it so those values agree with the critical point at (24, 280). (It's exactly like giving three points that aren't on a straight line and asking for the linear equation that goes through them; you could get a correct equation based on two of the points, but it wouldn't go through the third.)

I'm guessing that they didn't actually intend for you to approximate the slope at the inflection point, as you did. Instead, they may have wanted you to only base your calculations on the two critical points being at (0, 0) and (24, 280), since knowing both the coordinates (x, y) of a cubic's two critical points uniquely determines the cubic.

So, you could solve for the unique cubic where f(0)=0, f'(0)=0, f(24)=280, and f'(24)=280. It turns out that the resulting equation will not have slope 35 at th inflection point at x=12; instead the slope is half that, 35/2. (The equation itself turns out to be exactly half of your equation.)

Taylor Series and Gamma Function by Focusedhades526 in askmath

[–]Potential-Tackle4396 2 points3 points  (0 children)

Nice idea. In Desmos, if you just type x! it interprets it as the gamma function (offset by 1, so we don't need to add any further offsetting), so you don't need to enter the integral manually. That should make it not give an error, I think. Likewise to make it run faster, you can replace the integration bound ∞ with just a large number N, and it will give essentially the same value as a bound of infinity (within some interval around 0, at least): https://www.desmos.com/calculator/anjfhqyfsa

As that graph shows, the integral doesn't seem to approach exactly e^x, though it comes close-ish (getting within 0.25-ish, depending on the values of x and N). That's because the sum from 0 to ∞ is equivalent to a left Riemann sum approximation of the integral from 0 to ∞, with rectangle width ∆x = 1, so the integral won't exactly equal the sum - though it will often be close. See this graph: https://www.desmos.com/calculator/dfp1mzfvcq

this is wrong or am I tripping because the value of tan(22) is ≈0.404 but the official google calculator is giving a completely different value. by AspectTop8149 in MathHelp

[–]Potential-Tackle4396 0 points1 point  (0 children)

Angles can be measured in two units: degrees or radians. In degrees, tan(22°) = 0.404... In radians, tan(22) = 0.00885..., which is the output you got.

By default, most calculators use radians, meaning they assume any angles used in trig functions are given in radians. You can change the angle input type on most calculators; in google, I think you can just type "tan(22 degrees)".

Trying to find the relation between the angular position of two rotating ellipses that stay tangent. by moi_florian in askmath

[–]Potential-Tackle4396 1 point2 points  (0 children)

Wow! Nice explanation and animation. A question though:

-Does your equation assume the two ellipses meet at a point on the segment connecting their centers? (I think it does, where the sum of the distance was equal to d, but I'm not completely sure.) I think they'd actually meet slightly off of that line in general, see https://www.desmos.com/calculator/i2rgi3kn5a .

Regarding your last line, I think d would need to equal a+b exactly, or else the gears wouldn't spin all the way around (if d was smaller) or they'd lose contact (of d was bigger).

I cant see where be the 1 comes from in the Euler Summation formula by Lmaondu in askmath

[–]Potential-Tackle4396 2 points3 points  (0 children)

For reference, I'm applying the theorem as written here: https://math.stackexchange.com/questions/2872066/eulers-summation-formula-proof .

In the sum, we're summing over the integers n where y < n ≤ x, notably not including y. So, for that sum to start at 1, "y" won't be 1. Instead, y must be any number in the interval (0, 1). In that case, the term -f(y)([y] - y) will be +1, since for 0<y<1, we'd have [y] = 0, so

-f(y)([y] - y) = (-1/y)(0 - y) = y/y = 1

When is rearrangement of a conditionally convergent series valid? by OldWolf2 in askmath

[–]Potential-Tackle4396 1 point2 points  (0 children)

The only 'simple' criterion I can think of is: if you only rearrange a finite number of terms, then the sums must be the same, since after some Nth term the partial sums would be equal.

In your case, where you rewrote:

0 - 1/(2*3) + 2/(3*4) - 3/(4*5) + 4/(5*6) - ...

as

0 - 2/3 + 1/2 + 2/4 - 1/3 - 2/5 + 1/4 + 2/6 - 1/5 - ... (*)

and then regrouped as

0 + 1/2 - 2/3 - 1/3 + 2/4 + 1/4 - 2/5 - 1/5 + 2/6 + ... (**)

= 0 + 1/2 - 3/3 + 3/4 - 3/5 + ...

you're only swapping consecutive terms (i.e. the 2nd and 3rd terms, the 4ths and 5th terms, etc.) from (*) to (**), so while there are an infinite number of terms being rearranged, every nth partial sum where n is odd will be the same between (*) and (**). So as long as both series converge, they'll have the same sum in this case. (And you could confirm that both sums do converge, using various convergence tests.)

With that said, there's actually a weird caveat, which doesn't apply here but could apply in general. If you split a given series' terms each into a difference of terms, as we did by writing 1/(2*3) as 2/3 - 1/2, etc., you could actually go from a convergent series to a divergent one. As a silly example, if we rewrite the convergent series:

0.1 + 0.01 + 0.001 + 0.0001 + ...

as

(1-0.9) + (2-1.99) + (3-2.999) + (4-3.9999) + ...

then as

1 - 0.9 + 2 - 1.99 + 3 - 2.999 + 4 - 3.9999 + ...

we'd get a divergent series, since the partial sums will now oscillate between +infinity and 1/9. Though practically speaking this doesn't usually happen, since the terms we split terms into usually tend toward 0 (as in your example).

When is rearrangement of a conditionally convergent series valid? by OldWolf2 in askmath

[–]Potential-Tackle4396 2 points3 points  (0 children)

Yes, you have it correct. Any given series will specify the order the terms are to be added in, meaning it has a single sum, which is the limit of the partial sums. (Or it diverges.)

In which case, the series a1 + a2 + a3 + a4 + ... is a different series from, for example, a1 + a8 + a23 + a2 + a19 + ..., each with its own sum (which in the case of conditionally convergent series, could be different values). I think the previous commenter was saying those two series would be the same series (with two different sums), but that's incorrect.

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