A radical equation.

We solve a nice radical equation, using Descartes Rule of signs.

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[zanti]

Putting "qed" at the end of proofs gets boring. I'm going to start putting "atagpts".

[chasemarangu]

Nobody:
Michael Penn: spend weekend trying to solve 70th order polynomial

[ДенисЛогвинов-зе]

I haven't heard about descartes rule of signs, something new to me) Interesting staff)

[Agony]

You can try to prove that the function f(t) is monotonically increasing between zero and one. If so, it’s easy to illustrate that there is only one root between 0 and 1.

[cetjberg]

Good Job!

I once said to a mathematician "if you use Descartes' rule of sign . . ". He said "What's That?".

[UnderwaterCascade]

Can't we bypass most of the discussion? t^105-t^70-t^42+t^30 has 2 sign changes, so it has 0 or 2 positive roots, since 1 is a simple root, there must be exactly 2 positive roots. Together with 0, this gives 3 real roots to the original equation.

[flux]

If there's a 100k video soon (hopefully) can you show us some of your solutions that you do on paper with all your scratch work and failed attempts etc before you present them on a board, love the content keep it up, easily the best math youtuber on the site 👍

[philippenachtergal]

8:37 Hum, if you don't factor out t^40, you don't even need to do a smart pairing. Each term on the left part will be bigger than each term on the right part (when t>1) and there are more terms on the left so the left part is bigger.

[cernejr]

According to Wolfram Alpha: t≈0.979047; x≈0.0117155 . The graph of the function of t is very spiky due to the high powers.

[qqpit]

At 3:35, how does it follow that t is nonnegative? A negative value of t corresponds to a positive value of x since x = t ^ an even power.

[bobajaj]

i'm not sure about t being non negative as the power 210 is even so t can also be negative.

[tomatrix]

Decartes Law is something new for me aswell, very nice practical usage

[MarcoMate]

At 3:50, it should be underlined that if t is real, then sqrt (t^210) = |t|^105, because both sides of this equation must be positive. Thus we have two cases:
1) sqrt (t^210) = t^105 if t >= 0;
2) sqrt (t^210) = - t^105 if t <= 0.
These two cases bring to two different polynomial equations, which are:
1) t^30 - t^42 = t^70 - t^105 if t >= 0;
2) t^30 - t^42 = t^70 + t^105 if t <= 0.
These two equations BOTH have to be analyzed, not only the first equation as stated in this video. Each of these two equations could bring solutions t; in particular, the second equation could bring negative solutions t which are completely fine, due to the fact that t^210 = x is always positive. There could exist a solution x to the original equation which is positive and comes from a negative t, solution of equation 2), such that t^210 = x.
Indeed, a funny thing happens: the set of solutions of equation 1) with t positive contains exactly the opposite values of the solutions of equation 2) with t negative. What I'm saying is that if t is positive and solves 1), then -t is negative and solves equation 2), and vice versa. Thus, no new solutions x really come from equation 2), they all can be found from equation 1).

[GDPlainA]

0:56 imagine if he says "that's a good place to stop"

[goodplacetostop]

13:33

Not really homework but still an interesting question I’ve found on Maths Stack Exchange. Link to the topic after the problem.






A person is thinking of a number between 1 and 1000. What is the least number of yes/no questions that we can ask and know what that person's number is given that the person is allowed to lie on at most one of her answers.

[jamirimaj]

Why is t only bigger than or equal to zero? Since you're raising it to an even power (210 in this case), it should not matter whether t is positive or negative, right?

[InfolineIo]

9:45 why are there no roots when t>1?

[alexh]

So far the best problem you have solved! I really like your explanations.

[chrstfer]

How would one solve this for floating point numbers? Those are also a field, right? Just occurred to me that maybe it would be interesting to compare

[kevinmartin]

3:40 It isn't entirely obvious to me that we need x>=0. Although we want real x, that doesn't eliminate the possibility that some of the roots involved could be the complex non-principal roots (in addition to the square root which would be pure imaginary) and all the imaginary parts cancel out.

Perhaps there is an understanding that the other three roots are the principal root only.