Glad you corrected yourself, when you said"right" and instead you said" isn't it"

@davidwright8432

6 жыл бұрын

Hmm... I'd argue that 'Right' and 'Isn't it so?' are the same thing - and 'Isn't it?' is just an abbreviated form of 'Isn't it so?' 'True' works as well as 'so', - isn't it so? 'Isn't it' is idiomatic in some forms of English - just not N American! So no 'correction' needed!

@PerfectlyCrafted

6 жыл бұрын

david wright I liked how he corrected himself because "isn't it" Is like his catch phrase

@Somperzu

3 жыл бұрын

“Right” was the correct grammar, but “isn’t it” is funnier haha.

hahaha, you need to release an “isn’t it?” shirt!

@blackpenredpen

6 жыл бұрын

I hope I can find a way to do that soon. I would love to wear one myself hehehe

@JoJoJet100

6 жыл бұрын

i would so buy that, it would be the most niche shirt I own

Right. I mean "Isn't it?" Man, I started something with the unsuscribing and resubscribing...

@blackpenredpen

6 жыл бұрын

AndDiracisHisProphet LOLLLLLLLL

@blackpenredpen

6 жыл бұрын

AndDiracisHisProphet thanks man, you always brighten my days and also nights

@AndDiracisHisProphet

6 жыл бұрын

I'll do my very best

@blackpenredpen

6 жыл бұрын

AndDiracisHisProphet yay! Btw, I will answer your question here. I have been working on that channel for my school and I think I will put all the practice and solutions videos there.

@AndDiracisHisProphet

6 жыл бұрын

So the videos on this channel are more for your hobby and the others for your work?

Hey BlackPenRedPen I go to UC Berkeley and recently one of my friends told me that you are a professor on the campus, if one day I would love to meet you! I'm an undergraduate with the college of Chemistry and I"m such a big fan, and would love to meet you to ask you about great math textbooks and I just have so many questions I have for you

I'll definitely recommend that other channel to my students, and also I might be helping on translating it. :)

Another challenge Prove that Sin(2π/7) + Sin(4π/7) + Sin(8π/7) = (sqrt(7))/2

@angelmendez-rivera351

5 жыл бұрын

Aditya Vijj Use the multiple angle formula, then solve the polynomial equation which results, letting x = sin π/7.

@angelmendez-rivera351

4 жыл бұрын

sin(2π/7) + sin(4π/7) + sin(8π/7) = sin(3π/7) + sin(2π/7) - sin(π/7), since sin(4π/7) = sin(3π/7) by symmetry, and sin(8π/7) = sin(π + π/7) = -sin(π/7). sin(3π/7) = 3·cos(π/7)^2·sin(π/7) - sin(π/7)^3 & sin(2π/7) = 2·sin(π/7)·cos(π/7). Let a = sin(π/7) & b = cos(π/7). Then our desired expression is 3ab^2 + 2ab - a - a^3 = 2a(2b^2 + b - 1). Consider that sin(4π/7) = sin(3π/7) = 3ab^2 - a^3 = a(4b^2 - 1) = 4ab(2b^2 - 1), by using the quadruple angle formula. Rearranging, this gives us the polynomial equation 8b^3 = 4b^2 + 4b - 1. This equation is important. Let our desired expression be equal to x. Then x^2 = 4a^2·(2b^2 + b - 1)^2 = 4(1 - b^2)(2b^2 + b - 1)^2. (2b^2 + b - 1)^2 = 4b^4 + b^2 + 1 + 4b^3 - 4b^2 - 2b = 4b^4 + 4b^3 - 3b^2 - 2b + 1. Given the polynomial equation above, we can multiply by 2b to obtain 16b^4 = 8b^3 + 8b^2 - 2b = 4b^2 + 4b - 1 + 8b^2 -2b = 12b^2 + 2b - 1. We want 4(2b^2 - b - 1)^2 = 16b^4 + 16b^3 - 12b^2 - 8b + 4 = 12b^2 + 2b - 1 + 16b^3 - 12b^2 - 8b + 4 = 16b^3 - 6b + 3 = 8b^2 + 8b - 2 - 6b + 3 = 8b^2 + 2b + 1. Then, x^2 = (1 - b^2)(8b^2 + 2b + 1) = 8b^2 + 2b + 1 - 8b^4 - 2b^3 - b^2 = 7b^2 + 2b + 1 - 8b^4 - 2b^3. Multiply by 4 to get 4x^2 = 28b^2 + 8b + 4 - 24b^2 - 4b + 2 - 8b^3 = 4b^2 + 4b + 6 - 8b^3 = 7. This is the key discovery. 4x^2 = 7, so x^2 = 7/4, hence x = -sqrt(7)/2, or x = +sqrt(7)/2. Which of the two roots is x equal to? sin(x) is a monotonically increasing function on the interval (-π/2, +π/2), in which π/7, 2π/7, and 3π/7 lie. Therefore, sin(3π/7) > sin(2π/7) > sin(π/7) > 0, which implies sin(3π/7) + sin(2π/7) > 2·sin(2π/7) > sin(2π/7) + sin(π/7) > sin(2π/7), which implies sin(3π/7) + sin(2π/7) - sin(π/7) = x > 2·sin(2π/7) - sin(π/7) > sin(2π/7) > sin(2π/7) - sin(π/7) > 0, which implies x > 0, which implies x = sqrt(7)/2. Q. E. D.

@omarmagdyahmed4259

4 жыл бұрын

@@angelmendez-rivera351 shit

you are the only one that gave me the answer to my question when nobody did, i freaking love you

Your videos are excellent friend, I can see the passion you have for mathematics, keep it up. Greetings from Argentina.

@blackpenredpen

6 жыл бұрын

Thank you!!

That's a nice helpful rule. Thanks man.

Thank you for these videos. They're great for refreshing the ol' math center in my ol' noggin. ;)

Thank you..😊 have learnt a lot your the best

You're a bloody god, man. Thanks for helping me study for my uni entrance exam.

Hot damn, I've been hoping for more algebra videos to recommend to my algebra students!

@blackpenredpen

6 жыл бұрын

Alex Behlen yay!!! Thank you

thank you so much lol i missed the class when they taught us this and was struggling on the homework

keep up the great work dude :)

@blackpenredpen

6 жыл бұрын

pennyy thank you and thank you for your art work that you did for me!

I'm guessing you could come up with two cubic conjugates for it, by putting it on the complex plane and rotating by 2pi/3 (instead of rotating it by pi for the quadratic conjugate, which is by multiplying the second number by -1)

Thank you!

beautiful thing

As well as the derivation of (a+b)^3=a^3+b^3+3a^2b+3ab^2 or (a+b) (a^2+2ab+b^2)

Can you do the same thing with n-root and partial fraction??

Your videos are tremendously cool. It would be great if you created playlists for all math topics that are in your videos. (I know that you have playlists but they do not cover all of your videos) Hope you see this comment as an advice)

@blackpenredpen

6 жыл бұрын

Will try my best.

It was a pretty easy but in our algebra class we have to do rationalization of 1/cubrt(a)+cubrt(b)+cubrt(c) wich is pretty long

Can we solve in the complex world the integral from 0 to 1 of the logarithm with base i of x ? Thank you

@ribozyme2899

6 жыл бұрын

Well, for a start, the logarithm to base i of x is just ln(x) / ln(i), and ln(i) is i * pi / 2, so it's -2/pi * i * ln(x).

Thanks, great videos as usual... on the other note, maybe a new wireless shirt microphone would be a great upgrade! Just a thought

Here: F(n)=1/(x^(1/n) -y^(1/n)) G(n)={k=0 up to n}Sum(x^(n-k) *y^k) In general, F(n) can be rationalized by multiplying top and bottom by (G(n-1))^(1/n) if the power happens to be odd, then let z= -y, and you have a formula for 1/(x^(1/n) +y^(1/n))

@MrRyanroberson1

6 жыл бұрын

the cool thing is that this all came from the fact that for all x^n -y^n it generally factors first into (x-y)*G(n-1)

that was amazing

I am waiting for day when you will start teaching abstract math in same simple way you do.

@blackpenredpen

6 жыл бұрын

how abstract?

Thank you

3:56 Quick maffs

What if you have a root that is greater than 3?

I solved it before watching the video. You can also multiply the numerator and denominator by (a-b)²+ab, where a=cbrt(2) and b=cbrt(3)

@Alex-bc3tt

2 жыл бұрын

That is sure a long dry because -2ab will always add up with -ab to give -ab so you can just omit that extra unnecessary step

So, a quick question. If a^3+b^3 = (a+b)(a^2-ab+b^2), does this hold for all nonzero values of a and b? Or does it hold only for positive values of a and b?

@stephenbeck7222

6 жыл бұрын

Check it yourself. Choose, say, a = -2 and b = -3. Then on the left side we have ( (-2)^3 + (-3)^3 ) ( -8 + -27 ) = -35. On the right side, we have ( -2 + -3 ) ( (-2)^2 - (-2)(-3) + (-3)^2 ) = ( -5 ) ( 4 - 6 + 9 ) = ( -5 ) ( 7 ) = -35. Both sides work out to the same thing. So it works for at least one pair of negative values. Or you can just "foil" out the right side in the formula and verify that it simplifies to the left side, with no restrictions on any of the multiplying to account for positives or negatives. Or, if somebody was holding you captive and a condition of release was that you reproduce this formula but you completely forgot it (i.e., how did we come up with the formula in the first place), just do long division: (a^3 + b^3) / (a + b) and you should get a^2-ab+b^2 as the result.

@davidwright8432

6 жыл бұрын

Try some negative values for a and b, and see! All it takes is a single counterexample, and you know it wouldn't hold for negative numbers. Sometimes, even if you can't figure it out from first principles, you can fiddle around and see what happens. But you don't get proof - until an example fails. Then you'd know your choice was restricted to positive values.

@KnakuanaRka

5 жыл бұрын

Just multiply out the factors; you get a3-b3 without any steps that don’t work with negatives.

Could you please solve a quartic or a quintic equation? Just want to see how it's done.

@blackpenredpen

6 жыл бұрын

Taqee Mohammed I plan to do a cubic one first soon

@lewisbulled6764

6 жыл бұрын

There is no way to solve an equation with a fifth power as yet!

@Alex-bc3tt

2 жыл бұрын

@@lewisbulled6764 I assume you are a high school student because that is not true

@lewisbulled6764

2 жыл бұрын

@@Alex-bc3tt bro I wrote that 4 years ago 😂 I was back then, got much much more mathematical knowledge now lol

@Alex-bc3tt

2 жыл бұрын

@@lewisbulled6764 🤣🤣🤣🤣 I understand bro I once fought people here who said the square root of a negative number exists🤣 following year I went to varsity and learnt complex roots and comments came flooding attacking me 😫😫

Where is your other channel for Algebra?

Thanks❤

Hey! Could you calculate phi factorial?

@ffggddss

6 жыл бұрын

φ! = ∏(φ) = Γ(φ+1) I don't think there's a symbolic way to do that, like there is for half-integers, but my calculator says: Γ(φ+1) = 1.449229602269896600377879790629768... That's 33 decimal places. If you give this to the Wolfram Alpha website, it should be able to give you 50 places.

What are the uses of rationalizing a denominator? Or is this just a challenge for fun?

@gyroninjamodder

6 жыл бұрын

AreyouReggae The main reason is that pre digital computers it was hard to divide by irrational numbers accurately. It was much better to divide your irrational number by a rational number. It also serves to simplify some fractions and may make them easier to work with other fractions.

@ffggddss

6 жыл бұрын

"... hard to divide by irrational numbers accurately." - Exactly! I'm convinced this is how rationalizing denominators became an obsession in teaching & learning math. It made perfect sense when the best device you had to compute a numerical answer, was a slide rule. But nowadays, to get 1/√2, you just punch: 2, 1/x, √x into your calculator. And computational difficulty *is* still a good reason for rationalizing when you *don't* have access to a calculator. "... serves to simplify ... may make them easier to work with other fractions." - Yes, and sometimes, just the opposite! As a standalone, 1/√2 is manifestly, though only slightly, simpler than ½√2. Or 1/√57 vs (√57)/57

@blackpenredpen

6 жыл бұрын

AreyouReggae hi there. I have a video on that regard. kzread.info/dash/bejne/d5uNzqmRntSeqps.html Also, sometimes we can rationalize the denominator in integrals, limits or derivative to make the problems easier.

What about 1/sqrt(x)+cbrt(y). How would it be if the roots where different

@uchihamadara6024

6 жыл бұрын

Jacobo Zapata Perhaps you could write sqrt(x) as the sixth root of x^3, and write cbrt(y) as the sixth root of y^2. There is a formula to factor the sum of two sixth powers (infact, there is a way to write it for any nth power n>2) but I can't remember rn lol

@Jacob-uy8ox

6 жыл бұрын

Uchiha Madara that's sounds interesting, if you remember it put it on the feedback! :)

@uchihamadara6024

6 жыл бұрын

Jacobo Zapata Hey, good news, blackpen just uploaded a video on the exact question you asked. And he does it the way I mentioned. Correction on something I said earlier, there is a way to factor a sum and difference of any two numbers raised to odd powers, but for even powers you can only factor their difference (not their sum).

My doubt got solved Thanku

blackpenredpen, what's your name on brilliant.org and how can I send you questions?

Hmm, great little math problem.

Good video

Thanx dude😋😋😋😋

Sorry sir but I didn't understand it clearly but. Thank you so much for your effort

This is so easy. Just use sum of cubes formula. It should take long to figure out

How about, for example: Rationalize 2pi / (2 + pi)

@angelmendez-rivera351

5 жыл бұрын

Dawid Krainski You cannot rationalize transcendental numbers.

What if you have two different roots in the denominator, or three (square or cube, and both), is it possible to rationalize then?

@blackpenredpen

6 жыл бұрын

Yes and I will do a video on that soon

hello bprp! I need some help from you. I've watched your many videos about the function x^x in which you state that 0^0 is undefined. And we've just started with precalculus and limits at school, where my teacher told me that absolute zero^(absolute zero ) =1 but in case of limits, where we have approaching values of zero in the base and the exponent, it is an indeterminate form. Whereas, in your video, you state that 0^0 is undeFined and when you plotted the graph, you even left a hollow circle at zero. To add to my confusion, when I plotted the graph on demos (a graph plotting application) it included the point (0,1) and now I'm really confused regarding this. Please clear this man!

@davidwright8432

6 жыл бұрын

Let's look at the algebra situation. 0 = 0^1; no problem there. But if you try the usual technique for nonzero numbers - say q =q^1, then q^1/q^1 = q^(1-1) =q^0; but q/q = 1. so q^0 indeed = 1. BUT! if you try that one for 0, then what you're doing is forming '0/0', which is a no-no since division by zero is undefined. In the calculus case, you're approaching zero from above and below - but not quite ever getting there! You can get as close as you like, but not actually use the value 0. No graphing application has infinite visual resolution - nor does your eye! - so the more basic kinds of graphing application may try to sneak in a value for precisely 0. Naughty! It should really give some kind of warning. Like an open circle.

@suneetiyer81

6 жыл бұрын

david wright well... thanks a lot for your reply. Indeed, the graph plotter does give a hollow circle for undefined values (like y=sin(x)/x gives an open circle at x=0)but i didn't see any such hollow circles for f (x)=x^x and infact it even returned f (0)=1. As for your explanation for the limits, I'm totally Convinced. But could you please elaborate on why you say that the algebraic operation 0^0 is not defined and how you linked it to the 0÷0 form

Hello, how can we solve negative roots (not imaginary, example, x^(1/-2) understood?)? is there such a thing in math?

@creepedout7975

6 жыл бұрын

That would be equal to x^(-1/2), which is equal to 1/x^(1/2) So the result would be 1/sqrt(x)

@Rangsk

6 жыл бұрын

x^-y = 1/(x^y)

Now do it for 5th roots ;)

What about 1/(5root(2)+5root(3)+5root(5)+5root(7)+5root(11)+5root(13)) I have no idea how to go about this

@martinepstein9826

6 жыл бұрын

Let's call the denominator of that expression d. First you have to find the minimum polynomial of d over the rationals. That is, the lowest degree polynomial with rational coefficients and d as a root. Call this polynomial p(x) and call its constant term c. Note that (p(x) - c)/x is also a polynomial with rational coefficients and call this q(x). Now it's easy to see p(x) - q(x)*x = c. Plugging in d we get c = p(d) - q(d)*d = 0 - q(d)*d = -q(d)*d. So we can rationalize 1/d as -q(d)/(-q(d)*d) = -q(d)/c Note that in this problem p(x) is a 2^6 = 64th degree polynomial so BlackPenRedPen probably doesn't want to do this by hand.

Tqs to sir 🌹

Yay!

Why dont you answer in instagram?

Challenging You eliminate fifth root in binomial from denominator look like 1 /( 2 ^ 0.2 ± 3 ^ 0.2)

I thought this would be the conjugate: [2^(1/3)]^2 - [3^(1/3)]^2. Why is that wrong?

@mrocto329

2 жыл бұрын

Not wrong, but squaring cube roots doesn't help with rationalizing this fraction. As he said at the start of the video, we want to cube both cuberoots to get a rational number.

WOW!!!

but why do we rationalize them?

I need a help in rationalization . Will you reply me

merci

Plzzzzzzz do solve it cos25-sin25/cos25+sin25

I got exactly 1000th like 😘😉

This thing is rather easy lol...!

Hey Can u please solve this equation for me. Please 3x/2=X^(x^3/2)

@gabrielmello3293

6 жыл бұрын

Go study already

@AndDiracisHisProphet

6 жыл бұрын

That doesn't have an algebraically closed form. Are you sure you gave the right formula? www.wolframalpha.com/input/?i=3x%2F2%3Dx%5E(x%5E(3%2F2))

#rationalizing #rationalization #rationalize

Can you do (I!)? I am really wondering what is the anwpswer to that?

@tiagonewton4782

6 жыл бұрын

Do you mean "i" factorial? The imaginary unit?

Yesss i'm secant(video)=secant of video

(5x^2)+(y^2)+4xy+10x+6y+10=0 Russian mathematical olympiad 8 class

@Miaumiau3333

6 жыл бұрын

1, -5 (?)

@thesunisdark13

6 жыл бұрын

Rearrange to y^2+(4x+6)y+(5x^2+10x+10)=0. This is a quadratic in y, so for any value of x, for there to be a possible real value for y the discriminant (b^2-4ac) must be greater than or equal to zero. So (4x+6)^2-4(5x^2+10x+10)≥0. Expanding brackets and simplifying: (16x^2+48x+36)-(20x^2+40x+40)≥0. -4x^2+8x-4≥0. Multiply both sides by -1 (as -1 is negative, switch the inequality sign): 4x^2-8x+4≤0. Factorise to (2x-2)^2≤0. The square of a real number is never negative, and as the only solution to a^2=0 is a=0: 2x-2=0, so x=1 is the only possible value for x. Now substitute x=1 in: y^2+(4+6)y+(5+10+10)=0 y^2+10y+25=0 (y+5)^2=0 y=-5. So x=1, y=-5 is the only real solution.

@thesunisdark13

6 жыл бұрын

It can also be written as a sum of squares: (2x+y+3)^2+(x-1)^2=0, so both 2x+y+3 and x-1 must be zero, making x=1 and y=-5.

I could not solve it

Me: Just use Gröbner basis. *type some code into maxima* Me: Done.

Sir, your solution is correct in your point of view but not at all

10q🥰😍😍😍😍😍

Greetings from the Union of Soviet Socialist Republics.

@danilbutygin238

6 жыл бұрын

смешная шутка (нет)

@Mnemonic-X

6 жыл бұрын

Danil Butygin It is not funny. I am absolutely serious. Because the USSR still legally exists, but Russian Federation doesn't.

Correct! That's genius! I was also trying to rationalize a denominator with three square roots until I found a solution. It is done here: www.easycoursesportal.com/algebraicfractionscourse/courseb/Less-13.htm But what I've been trying to do recently is rationalize a denominator consisting of three cube roots. I've tried everything and I end up with an irrational denominator every time. So, is there any way to rationalize a denominator consisting of three cube roots? I know it could be very complicated but please try to do it.

@mostafakhaled9702

6 жыл бұрын

Zacharie Etienne Nice idea. But I've tried dividing a^3+b^3+c^3 by a+b+c by many ways and didn't come with any answer. Also, I see that he loved your reply. This means that either he found a solution or he is trying to find it. But please try your best. This time, this is the real challenge not like last time :D

@thesunisdark13

6 жыл бұрын

a+b+c isn't a factor of a^3+b^3+c^3, because if it was, then when a+b+c=0 a^3+b^3+c^3=0, which isn't always true - say a=2, b=-1, c=-1.

@mostafakhaled9702

6 жыл бұрын

Thomas Finn But I think there must be a way to rationalize three cube roots in the denominator. If there's a way to rationalize three square roots, there must be a way to rationalize three cube roots.

First

@sem5776

6 жыл бұрын

Third

You could do that or you could just not rationalize it because rationalizing is a pointless and dumb tradition that has yet to die out.

Ur accent is not clear

@blackpenredpen

6 жыл бұрын

Show us yours.