is it better if we connect the points, for example, (1/5,0) and (0,4/5) and so on? in this case, the intercept in x and y-axes add to 1, and there will be no error terms, all the straight lines will be tangent lines supporting the envelope. p.s., great video! and I definitely want to see the other parts on this topic! thanks! I also appreciate your approach of doing it from the "first principles" of calculus, without going straight to partial derivatives as in standard texts. great work!
Thanks for the message. I figured there must be a nicer way, but couldn't find it when searching around, so I went with what I knew and was delighted at the calculus I got to use. The construction I used (i.e. from (1/5,0) to (0,1), etc.) was the construction I remembered doing as a kid. But you are likely right if you connected (1/5,0) to (0,4/5) the numbers must be nicer and the derivative is probably better, but I wasn't clever enough to think about that. Do you happen to a derivation with partial derivatives? I'd like to look at that. And I will continue my analysis soon (maybe within the next week or so), my other work continued with my weird construction, but maybe I'll keep doing that as it will keep my results somewhat unique?
@@Dr.LovelessTeachesMath Oh, I see that the problem changes, then. The partial derivative solution is in the wiki article en.wikipedia.org/wiki/Envelope_(mathematics), but the top right picture there is somewhat misleading, be careful that later on in the article it mentions that the x-intercept and y-intercept add to 11 (the picture is only shows the 10x10 piece). Another nice exposition would be this #SoME1 video by @paralogical-dev ua-cam.com/video/fJWnA4j0_ho/v-deo.html I guess for fixed n, we get an envelope by y=(\sqrt{1/n+1}-\sqrt{x})^2 parameterized by n, and this sequence of envelopes will tend to y = x - 2 sqrt(x) + 1 as expected. Here's some MMA code n = 20; p1 = Plot[ Table[(1 - (n + 1)/k) x + (n + 1 - k)/n, {k, 1, n}], {x, 0, 1 + 1/n}, PlotRange -> {0, 1 + 1/n}, AspectRatio -> 1]; p2 = Plot[(Sqrt[1 + 1/n] - Sqrt[x])^2, {x, 0, 1 + 1/n}, PlotStyle -> Orange]; Show[p1, p2] ----------------- Yet another way of doing this, I guess, is to use the Titu's lemma en.wikipedia.org/wiki/Titu%27s_lemma, which is a variant of Cauchy-Schwarz. Consider the family of lines x/a+y/b=1 parameterized by a,b>0 s.t. a+b=1, then by Titu's lemma, 1=x/a+y/b=(x^(1/2))^2/a^1+(x^(1/2))^2/b^1>=(x^(1/2)+(y^(1/2))^2/(a+b)^1=(x^(1/2)+(y^(1/2))^2, and the equality can be attained, which is the envelope we want.
is it better if we connect the points, for example, (1/5,0) and (0,4/5) and so on? in this case, the intercept in x and y-axes add to 1, and there will be no error terms, all the straight lines will be tangent lines supporting the envelope.
p.s., great video! and I definitely want to see the other parts on this topic! thanks!
I also appreciate your approach of doing it from the "first principles" of calculus, without going straight to partial derivatives as in standard texts. great work!
Thanks for the message. I figured there must be a nicer way, but couldn't find it when searching around, so I went with what I knew and was delighted at the calculus I got to use. The construction I used (i.e. from (1/5,0) to (0,1), etc.) was the construction I remembered doing as a kid. But you are likely right if you connected (1/5,0) to (0,4/5) the numbers must be nicer and the derivative is probably better, but I wasn't clever enough to think about that. Do you happen to a derivation with partial derivatives? I'd like to look at that. And I will continue my analysis soon (maybe within the next week or so), my other work continued with my weird construction, but maybe I'll keep doing that as it will keep my results somewhat unique?
@@Dr.LovelessTeachesMath Oh, I see that the problem changes, then.
The partial derivative solution is in the wiki article en.wikipedia.org/wiki/Envelope_(mathematics), but the top right picture there is somewhat misleading, be careful that later on in the article it mentions that the x-intercept and y-intercept add to 11 (the picture is only shows the 10x10 piece).
Another nice exposition would be this #SoME1 video by @paralogical-dev ua-cam.com/video/fJWnA4j0_ho/v-deo.html
I guess for fixed n, we get an envelope by y=(\sqrt{1/n+1}-\sqrt{x})^2 parameterized by n, and this sequence of envelopes will tend to y = x - 2 sqrt(x) + 1 as expected.
Here's some MMA code
n = 20;
p1 = Plot[
Table[(1 - (n + 1)/k) x + (n + 1 - k)/n, {k, 1, n}], {x, 0,
1 + 1/n}, PlotRange -> {0, 1 + 1/n}, AspectRatio -> 1];
p2 = Plot[(Sqrt[1 + 1/n] - Sqrt[x])^2, {x, 0, 1 + 1/n},
PlotStyle -> Orange];
Show[p1, p2]
-----------------
Yet another way of doing this, I guess, is to use the Titu's lemma en.wikipedia.org/wiki/Titu%27s_lemma, which is a variant of Cauchy-Schwarz.
Consider the family of lines x/a+y/b=1 parameterized by a,b>0 s.t. a+b=1, then by Titu's lemma,
1=x/a+y/b=(x^(1/2))^2/a^1+(x^(1/2))^2/b^1>=(x^(1/2)+(y^(1/2))^2/(a+b)^1=(x^(1/2)+(y^(1/2))^2,
and the equality can be attained, which is the envelope we want.
@@chaoshengzhe Thanks so much! Here is a video in response: ua-cam.com/video/RERZGaSZjdc/v-deo.html