Renter - page 9

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Integer:

I don't know, I have it written down what the formula is and all the variables are defined. Let me also clarify - this is the amount of profit taken each month (not the total profit for m months).

It remains to derive the formula for the sum of the series, you wrote that you do it easily - do it. Then take the derivative, equate it to zero...


In my notations, your formula for the current month's withdrawal looks like this: , for the amount in period t:, is exactly the same as the one I got above.

Consequently, breaking down the beast-like derivative of this function is as difficult as the one above.

I think you can try to pre-prologarithmize f and then look for its maximum... Maybe it will be easier that way.

avtomat:

And then, in the second step, open the valve that divides the flow into two parts. This will change the input flow.

You don't see the solution yet?

No, I don't know what you're thinking. Tell me.

 
Integer:

There are some that even the Pythagorean theorem, as interpreted by them, can't be understood.

OFFTOP:

At school they gave the most succinct proof of Pythagoras' theorem.

  1. A right-angled triangle is uniquely defined by the hypotenuse(c) and one acute angle(alpha).
  2. Therefore the area of a right triangle can always be expressed through the hypotenuse as follows: S = c^2 * f(alpha), where f is some function.
  3. In the figure, angles 1 and 2 are equal(alpha).
  4. The area of a large triangle is equal to the sum of areas of small triangle: S = S1 + S2, or from (2) so c^2 * f(alpha) = a^2 * f(alpha) + b^2 * f(alpha).
  5. From where we get c^2 = a^2 + b^2.

Note, the basic simplest (non-standard) idea is p.2. No knowledge of properties of similar triangles is used, also no knowledge of trigonometry is needed to understand the existence of function f either. I.e. such a proof can be given in primary schools after well (not as usual) explaining to children what area is.

 
hrenfx:

OFFTOP:

In school they gave the most succinct proof of Pythagoras' theorem.

In what grade?

The formula S = c^2 * f(alpha) is not obvious to a 7th grader. It is taking for granted that it kind of is.

 
Neutron:


Accordingly, breaking the beast-like derivative of this function is just as hard as the above one.

Is the whole process stuck with the derivative?

Is this function x0*k*(1-(1+q-k)^2)/(k-q)?

If so, it's like no problem, I've solved them easily, just need to remember a bit. The variable q?

 
sergeev:

in which class?

The formula S = c^2 * f(alpha) is not obvious to a 7th grader. It is taking for granted that it kind of is.

Almost any child who has been introduced to the concept of the area of a figure well enough to feel it will have little difficulty understanding the above proof.

If a child truly understands what area is, he understands the measure of it and also understands that the area of any figure can be expressed through its characteristics (in this case the hypotenuse and angle) that uniquely define the figure.

No knowledge of the properties of similar triangles and trigonometry is needed.

 

I was on a visit recently and saw two stone pyramids (similar to Egyptian pyramids). Took them in my hands and put them at their bases (they are slightly different in size):

And came up with another proof of Pythagoras' theorem (clear from the construction).


Integer:
Весь процесс уперся в производную?
Вот эта функция - x0*k*(1-(1+q-k)^2)/(k-q)?
Если это так, то это как бы не проблема, я их легко решал, только вспомнить надо немного. Переменная q?


No, the problem is the derivative of k from:

It has to be equated to zero and solved with respect to k.

 

I can't do it the smart way, so I'll make it simple:


 Shoot as much as possible each time
At the end of the period

10 000 5,00% 3,00%
10 000 5,00% 3,00%
1 10 200 500 300
10 500 500
2 10 404 510 306
11 025 525
3 10 612 520 312
11 576 551
4 10 824 531 318
12 155 579
5 11 041 541 325
12 763 608
6 11 262 552 331
13 401 638
7 11 487 563 338
14 071 670
8 11 717 574 345
14 775 704
9 11 951 586 351
15 513 739
10 12 190 598 359
16 289 776
11 12 434 609 366
17 103 814
12 12 682 622 373
17 445 855 513



4 024


513









=B12+C13-D13 =B12*$C$1 =B12*$D$1
=F12+G13-H13 =F12*$G$1 =F12*$H$1

Let's say there is 10,000 on the deposit at the beginning of the period. Each period we add 5% to the deposit and reinvest them to the deposit. Each period we are allowed to withdraw only 3%.

If you withdraw all 3% of your money each period, all we get more than 4k$ (and do not give a shit about the deposit), otherwise we get only 0.5k$ (but with much on the deposit).

 
hrenfx:

Almost any child who has been introduced to the concept of the area of a figure well enough to feel it will have little difficulty understanding the above proof.

If a child truly understands what area is, he or she understands the measure of it and also understands that the area of any figure can be expressed through its characteristics (in this case the hypotenuse and the angle), which uniquely define the figure.

That's the point, all of the above is "it feels like this". That "it can somehow be expressed through something".

But it's not a rigorous proof.
 
Rich:

I can't do it the smart way, so I'll make it simple:


 Shoot as much as possible each time
At the end of the period

10 000 5,00% 3,00%
10 000 5,00% 3,00%
1 10 200 500 300
10 500 500
2 10 404 510 306
11 025 525
3 10 612 520 312
11 576 551
4 10 824 531 318
12 155 579


That's why we need a general analytical solution, not to draw such tables, but to substitute two input values into a simple formula and get the answer.
 
sergeev:
That's the point, all the above is "it feels like that's how it's going to be". That "it can somehow be expressed through something".

But it's not a rigorous proof.

What kind of hard evidence is that?! It's obvious:

  1. A right-angled triangle is uniquely given by the hypotenuse and the acute angle - obvious.
  2. So the area (perimeter and any other characteristics) of a right triangle is uniquely expressed by the hypotenuse and the angle - obvious.
  3. The measure of area is a square. So it follows from (2) that S ~ c^2, and since the angle to the hypotenuse uniquely defines the triangle, then S = c^2 * some dimensionless relation(f) to the angle(alpha) - obviously.
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