Basic Mathematics - 2021-01-17

« first day (985 days earlier) ← previous day next day → last day (1194 days later) »

3:23 AM

A: Should my 8th graders see a proof of the Pythagorean Theorem?

Actually the most 'intuitive' proof of the Pythagoras theorem involves cutting the big square into only 3 pieces and reassembling them into the two smaller squares (using merely translations). Absolutely no subtraction is involved! (See the last section below for why subtraction of area is a nont...

7 hours later…

RaphX

10:52 AM

Hello @user21820

I need some help regarding a binomial theorem problem

user102532

When I say $ z = r(\cos \theta + i \sin \theta)$

There are two ways we can find it’s square value ,

$z^2 = r^2(x^2 + y^2) = r^2(\cos \theta^2 + i \sin \theta^2)$ (Here I did the $x^2$ and $y^2$ separately)

Ok but if we directly square $= r(\cos \theta + i \sin \theta)$ this

Then , should we also get $2\sin\theta \cos \theta$ which means $ r^2(\cos \theta + \sin\theta)^2$

Why not in 1st one because when x = r cos theta , we will only get r ^2 = cos ^2 theta .

RaphX

I have to find the coefficient of x^10 in (x + x^2 + x^4 + x^8 + ...........) ^ 4

I tried converting this into a simpler form with the help of infinite GP formula after which I am getting x ^ 4 / (x ^ 2 - 1) ^ 4

can you help me in this? @user21820

anonymous_user

11:47 AM

@user21820 Just dropping in to say "Hello!" for now. Unfortunately, my schedule is a bit erratic, so I may be an infrequent visitor here. But, I will be happy to share in discussions with you whenever possible. :)

user21820

12:07 PM

@anonymous_user Sure! Look forward to interesting questions from you! =)

RaphX

12:36 PM

@user21820 ?? any idea on how to proceed?

user21820

@RaphX Hold on; busy with something else.

RaphX

okk , sorry I thought you were free

1 hour later…

user21820

1:45 PM

@user102532 You wrote "cos θ^2". That means "cos(θ^2)" and so is certainly wrong.

If you want the cos to be applied before the square, you have to write "cos(θ)^2".

If you fix that, then the first way is still wrong because y^2 is not i·sin(θ)^2.

In any case, "z^2 = r^2(x^2 + y^2)" is certainly wrong and I have absolutely no idea where you got that from...

Your second way is also wrong because you did not even square it directly. Write it out properly and carefully manipulate the result algebraically using only the basic field properties and the identity i^2 = −1.

@RaphX That is fine, but you still have to expand the (x^2−1)^4 before you can convert it back.

That is, you still need long-division to find 1/(1−4x^2+6x^4−4x^6+1) and then find the appropriate coefficient.

Hang on, I didn't realize you started with a non-GP!

You have powers of 2 in the exponent! In any case, it is almost always easier to use plain combinatorics and never generating functions. You want 4 powers of 2 to add to n = 10. There aren't many possibilities, and you can even find a formula for general n.

RaphX

could you explain more? @user21820

user21820

2:33 PM

@RaphX Well at least you have to see for yourself that it is not a GP.

And then simply find the answer by thinking about what it really means instead of looking at the power series. Just express the problem in combinatorial terms rather than in terms of coefficients.

RaphX

2:52 PM

ok yeah I did a mistake there

Now I know that ( 2 + 2 + 2 + 4) = 10 or (1 + 1 + 4 + 4) = 10

@user21820 am I going in the correct way?

user21820

3:16 PM

@RaphX Yes.

RaphX

3:42 PM

ok for the first case we can get 4C1 * 4C3 = 4 ways in total and for the second case we can get 4C2 * 4C2 = 36 ways, is this ok? @user21820

user21820

3:59 PM

@RaphX I'm not sure what is the point of using binomial coefficients for such small n.

If you want, try to solve for general n, then I can check whether your reasoning is correct.

That is, 4 powers of 2 that sum to n. How many ways?

RaphX

okk I got it now so for the 1st case where I take three 2s and one 4 we can have 4 in any 4 places giving us a total of 4 for that and for the 2nd case we can have 4C2 ways of arranging either the 1s or the 4s after which the remaining would fix themselves thereby giving us 6 as the total

So all together I would get (4 + 6) = 10 ways right? @user21820

user21820

4:23 PM

@RaphX Yes, and I still want to see how you would solve the general n. =)

DavidP

4:59 PM

I have also a combinatorics question: Is there a nice double counting argument for the identity 2\cdot\sum_{k=0}^n (n-k)\binom{2n}{k}=n\binom{2n}{n}? The right-hand side counts something like choose a team from n person out of n men and n woman and then choose a captain. Unsure about the left-hand side.