Engineer's answer: √2 - 1 is almost 0.5. So, (√2 - 1)¹² is almost zero 😂.
@kennethgee2004
16 күн бұрын
would say though that 0.5 is 1/2 which is 2^-1, so the answer is approximately (2^-1)^12. We can calculate that using powers of 2 fairly easily, so that is also approximately (2^12)^-1. This might be more what an engineer would say, but I get the joke and surprisingly they are not wrong.
@oahuhawaii2141
13 күн бұрын
@kennethgee2004: But a good engineer wouldn't say that because 2⁻¹² is an order of magnitude off from (√2 - 1)¹² .
@davidbrisbane7206
13 күн бұрын
@@oahuhawaii2141 🤣😂🤣🤣🤣😂👍
@Sergey_Moskvichev
11 күн бұрын
Ответ любителя: (√2-1)¹² более красиво и кратко выглядит, чем 19601-13860√2. 😊
@maherhaddad6455
9 күн бұрын
And it's true
@oahuhawaii2141
13 күн бұрын
The fast, direct, straightforward method: (√2 - 1)¹² = [[ (√2 - 1)²*(√2 - 1) ]²]² = [[ (3 - 2*√2)*(√2 - 1) ]²]² = [[ 5*√2 - 7 ]²]² = [ 99 - 70*√2 ]² = 9801 + 9800 - 140*(100 - 1)*√2 = 19601 - 13860*√2 The slow, complicated, error-prone method: Let x = √2 - 1 . We find: (x + 1)² = 2 x² + 2*x + 1 = 2 x² = 1 - 2*x . x¹² = [[ (x²)*x ]²]² = [[ (1 - 2*x)*x ]²]² = [[ x - 2*x² ]²]² = [[ 5*x - 2 ]²]² = [ 25*x² - 20*x + 4 ]² = [ 29 - 70*x ]² = 841 + 4900*x² - 4060*x = 5741 - 13860*x = 19601 - 13860*√2 This is faster than what was done in the video. But both evaluate to the same value of about 0.0000255089026236... At 15:00, you make the mistake of saying 0.000025508 is the exact value.
@Bisinski
8 күн бұрын
Im starting to relearn maths,.can you explain more in detail how the 12 power was fractioned in the second step
@kateknowles8055
8 күн бұрын
@@Bisinski Oahu worked the square out, then used that to work the cube out. (3- 2sqrt(2)) * (sqrt(2)-1)= 5sqrt2- 7= approx 0.07106 Then squared the cube to get the sixth power. Lastly squaring the sixth power to get the required twelfth power. I recommend finding the calculator on your phone or computer and increasing skill with that as well as with the numeracy and algebra. Enjoy your persistence as well as the corrct result
@aspenrebel
7 күн бұрын
I like ur first way better
@MattColler
4 күн бұрын
I did the same - I'll take squaring a binomial (generating three terms) any day over multiplying two different binomials (generating four terms). You can save even more work on the first two steps using the identity: (a - b)³ = a³ - 3a²b + 3ab² - b³ So: (√2-1)³ = (√2)³ - 3.(√2)² + 3.√2 - 1 = 2√2 - 6 + 3√2 - 1 = 5√2 - 7
@xl000
16 күн бұрын
This is what non math people think what hard maths are.
@YAWTon
16 күн бұрын
He seems to think that it is higher mathematics, and that his "trick" is so good that he can bring the (essentially) same problem again and again.... (today, yesterday, 1 month ago, 7 months ago, 11 months ago).
@nonickname142
11 күн бұрын
so agreed
@aspenrebel
7 күн бұрын
Hard math is.
@xl000
7 күн бұрын
@@aspenrebel I would love to see abstract algebra on this channel
@aspenrebel
7 күн бұрын
@@xl000 pi r squared. No! Pie are round.
@Vega1447
16 күн бұрын
Exactly. This is turning a straight forward piece of arithmetic into click bait.
Does anyone else find these videos really tedious cos of the amount of unnecessary duplication? Why write out an entire complicated expression again merely to move a digit from one side of the equals sign to the other and change the sign or to subtract a "1" from both sides or why write out "(x^2)^2" in an entire duplicated line with the only difference being that it's then written as "x^2.x^2" and then a further duplication of the entire exact same expression merely to finally express it as "x^4"? Please have mercy on us, for goodness sake.
@todd8155
6 күн бұрын
Double tap on the right side of the video to move forward by 5 second intervals, or press the right arrow. The steps are there if you need them...
@debrainwasher
16 күн бұрын
There is a much simpler way by raising the sqrt(2)-1 to the 2nd, 2nd and 3rd power, since 2·2·3=12. this results in two times binomial (a-b)² and one (a-b)³. Done.
@user-nr6tt5he7g
15 күн бұрын
I'd rather start with 3, then twice 2. But the idea is the same. (V2-1)^3 = (V2)^3 - 3(V2)^2 + 3V2 - 1 = 2V2 - 6 + 3V2 - 1 = 5V2 - 7 (V2-1)^6 = (5V2-7)^2 = 25(V2)^2 - 2*7*5V2 + 49 = 50 - 70V2 + 49 = -70V2 + 99 (V2-1)^12 = (-70V2+99)^2 = 4900*(V2)^2 - 2*99*70V2 + 99^2 = 9800 - 13860V2 + 9801 = -13860V2 + 19601
@EnginAtik
14 күн бұрын
You may not even even use the cubic expansion 2.2.(2+1) = 12. (17-12*sqrt(2))^2*(17-12*sqrt(2))=19601-13860*sqrt(2)
@oahuhawaii2141
13 күн бұрын
@EnginAtik: Well, I can't cube (17 - 12*√2) in my head! It's much easier to square for x², multiply for x³ = x*x², and square twice for x⁶ and then x¹² . The coefficients will be smaller when the "cubing" is done early; squaring isn't as bad. (((√2 - 1)³)²)² { cube it in my head } = ((5*√2 - 7)²)² { square it in my head } = (99 - 70*√2)² { must put down partial results } = 9801 + 9800 - (100-1)*140*√2 { manageable } = 19601 - 13860*√2
@si3986si
12 күн бұрын
I was thinking the same just break the power into 2.2.3 and apply formula I got the same ans ❤👍
@crescentsg090877
9 күн бұрын
This is too long, (2^o.5 - 1)^2 = 3-2*2^o.5 For another square = 17-12*2^o.5 Using cubic formula= 17^3 + 3*17*12^2*2 - (3*17^2*12+12^3*2) 2^o.5 Alternatively [sin 5pi()/24 * cos pi()/12]^12
@user-xr2cn3vx1u
16 күн бұрын
Calculate {0,1}.{{-1,2},{1,-1}}^12. Diagonalizate the matrix.
It seems that your answer would give an negative solution? Not possible in x¹².
@user-nr6tt5he7g
15 күн бұрын
You have a mistake in x^12, it should be x^12 = (x^6)^2 = 4900x^2 - 2*29*70x + 841 = -9800x + 4900 - 4060x + 841 = -13860x + 5741 = -13860V2 + 19601
@okohsamuel314
15 күн бұрын
@@user-nr6tt5he7g ... Edited, thanks!
@barneynisbet6267
16 күн бұрын
Surely anyone interviewing at Harvard for Maths has knowledge of the binomial theorem? It’s a trivial solution.
@oahuhawaii2141
13 күн бұрын
Well, if you can do error-free work for (x + y)¹² , then go for it! Most folks will make an error or two in the process.
@agrushnev
15 күн бұрын
and hence in its turn we obtain a rational approximation for √2 ~ 19601/13860
@exoplanet11
Сағат бұрын
Cool! Somebody tell Pythagoras.
@navghtivs
Күн бұрын
How many of us are here just to see how badly this guy fumbles an easy problem?
@saleemalkoury5572
16 күн бұрын
Ohh dear, you can easily reduce it to 17 -12√2 to the power of 3 and then use newton binomial series which will be way faster
@oahuhawaii2141
13 күн бұрын
The exponent of 12 factors as 2*2*3. That can be permuted in 3 ways. We can see that squaring big binomials is easier than cubing them. The former likely can be done in my head, whereas the latter likely needs scratch paper. I opt to cube first with small numbers and then square twice. BTW, they're easier to compute when grouped in odd and even powers for the element with a square root: (x - y)² = (x² + y²) - 2*x*y (x - y)³ = x*(x² + 3*y²) - (3*x² + y²)*y The easiest way: (√2 - 1)¹² { 12 = 3*2*2 } = (((√2 - 1)³)²)² { Reference the cubic formula … } = ((5*√2 - 7)²)² { Square this in my head } = (99 - 70*√2)² { Next square isn't as easy ... } = ((100-1) - 70*√2)² { Make it manageable } = (100-1)² + 70²*2 - 2*(100-1)*70*√2 = 19601 - 13860*√2 ≈ 0.0000255089026236... { Calculator verified } The harder way: (√2 - 1)¹² { 12 = 2*3*2 } = (((√2 - 1)²)³)² { Square this in my head } = ((3 - 2*√2)³)² { Reference the cubic formula … } = (99 - 70*√2)² { Same as above } = ... = 19601 - 13860*√2 The hardest way: (√2 - 1)¹² { 12 = 2*2*3 } = (((√2 - 1)²)²)³ { Square this in my head } = ((3 - 2*√2)²)³ { Square this in my head } = (17 - 12*√2)³ { Reference the cubic formula … } = 17*(289 + 3*288) - (3*289 + 288)*12*√2 { Ugh! } = 17*(4*288 + 1) - (4*288 + 3)*12*√2 = 17*1153 - 1155*12*√2 = 11530 + 8071 - 4620*3√2 = 19601 - 13860*√2
@rogerphelps9939
16 күн бұрын
Just use Pascal's triandle to get the binomial coefficients for the sixth power say. Then it is straightforward to solve by squaring.
@oahuhawaii2141
13 күн бұрын
Better to cube, then square twice. And if the cube isn't easy, then square and then multiply.
@mateoclivio
3 сағат бұрын
It works and is faster
@RyanLewis-Johnson-wq6xs
14 күн бұрын
(Sqrt[2]-1)^12=19601 - 13860Sqrt[2]
@gennarofontanarosa2062
16 күн бұрын
Tedious, boring, long winded horsebleep. I hate long winded crap to solve an easy problem.
@user-by5oc6yt6d
3 күн бұрын
С таким же решением я предложу решить квадрат бесконечности...
@user-bz5jh6kw8b
2 күн бұрын
Квадрат бесконечности это бесконечность
@surindermadahar
19 сағат бұрын
This solution is probably as cumbersome as evaluating the expression using the binomial expansion.
@ToddKunz
6 күн бұрын
The thing I appreciate the most about your videos is that you go step by step and you don't skip any steps. Thank you for that.
@linsqopiring6816
15 күн бұрын
This was really a fun video for me, don't care what the math snobs in the comments say. I did alright in high school math decades ago and I can tell you the level was geared perfectly for me! Which means I would expect that to be the case for most adults since most adults have high school math. Also it had that beautiful quality you find in math when you start with a basic expression go through a long journey and in the end things finally simplify down to a basic expression again. Loved it.
@byronwatkins2565
5 күн бұрын
I would use the binomial theorem (i.e. Pascal's triangle).
@Paul_Hanson
15 күн бұрын
Actually, if you plug the original expression into your calculator and then plug in the derived formula you will not get the same answer because of rounding error. There is no such thing as an exact answer on a calculator when the expression involves an irrational number (and even most rational numbers have no exact representation on a calculator). So each method of calculating the answer when plugged into a calculator only comes up with an approximate answer. In this case the difference is significant if your calculator only has about 9 or 10 digits of precision. So which answer is closer to the truth, and why? If instead we let x=1/(sqrt(2)-1)=sqrt(2)+1 and use the method demonstrated in this video we get the formula: x^12=19601+13860sqrt(2) (sqrt(2)-1)^12=(1/x)^12=1/(19601+13860sqrt(2)) If you plug this formula into your calculator the answer should be a much better approximation than the formula derived in the video. Can you see why?
@oahuhawaii2141
13 күн бұрын
(19601 - 13860*√2) vs (19601 + 13860*√2)⁻¹ My calculator returns the same result, since it has great precision. However, it's always good to be aware of finite precision in calculations and alter the computation to avoid problems, such as subtracting 2 numbers that are very close to each other. That's why my old HP has the [eˣ - 1] function, which is targeted for x near 0.
@Paul_Hanson
12 күн бұрын
@@oahuhawaii2141 Possibly your calculator is doing 64 bit arithmetic but only displaying 9 or 10 digits. In that case the difference will be in the digits that aren't displayed. If you subtract one result from the other you will probably get a non-zero answer. The size of the difference should give you a clue as to how many digits of precision your calculator has.
lf looking for an approximate value you can do with much less arithmetic and numerical calculation. let a=sqrt(2) - 1 and b=-sqrt(2)-1 we have ab=-1 and a+b=-2 (a+b)^2=a^2+b^2+2ab. =>. a^2+b^2=6 (a^2+b^2)^2=a^4+b^4+2a^2b^2. => a^4+b^4=36-2=34 (a^4+b^4)^3=a^12+b^12+3a^4b^4(a^4+b^4) => a^12+b^12 = 34^3-3*34 = 34(34^2-3) ~ 39000 note that a
@kotrynasiskauskaite4995
7 күн бұрын
Taking the longer route I see. When you had x^2 could have substituted that into x^12 as 12=2*6 and have that expression to the power of 6. 6=2*3 so you just had to power that to 2 quite simply and then 3.
@mikec6347
2 күн бұрын
And that folks is how I met your mother. Any questions? 😎
@Utesfan100
14 күн бұрын
Repeated squarings are faster. X^2=3-2root(2) X^4=17-12root(2) X^8=577-408root(2) Now foiling the last two gives the answer.
@oahuhawaii2141
13 күн бұрын
But your last step is going to be hard, which is why you left it out. It's better to cube first, then square twice. And if the cube isn't easy, do the square and then multiply. (√2 - 1)¹² = (((√2 - 1)³)²)² { cube it in my head } = ((5*√2 - 7)²)² { square it in my head } = (99 - 70*√2)² { must put down partial results } = 9801 + 9800 - (100-1)*140*√2 { manageable } = 19601 - 13860*√2
@herbie_the_hillbillie_goat
16 күн бұрын
There is a really simple way to calculate coefficients of the binomial theorem as you go without resorting to drawing Pascal's triangle.
@oahuhawaii2141
13 күн бұрын
Yes, demonstrate this for the 12th power: (x + y)¹² = ???
@herbie_the_hillbillie_goat
12 күн бұрын
@@oahuhawaii2141 Sure, I'd be happy to explain! Bear with me, since KZitem comments aren't the best for math. 1. *Binomial expansions are symmetric*, so we only need to figure out half the coefficients. 2. *Exponents of x decrease by 1* each term, while *exponents of y increase by 1*. 3. The *first coefficient is always 1*. Let's start with x^12. The next term will have the form Cx^11y. - Multiply the coefficient and exponent of x in the previous term: 1 * 12 = 12. - Divide by the exponent of y in the current term: 12 / 1 = 12. So the second term is 12x^11y. For the third term Cx^10y^2: - Multiply the previous coefficient by the exponent of x: 12 * 11 = 132. - Divide by the exponent of y: 132 / 2 = 66. Now you have 1x^12 + 12x^11y + 66x^10y^2. Keep going: - 66 * 10 / 3 = 220 - 220 * 9 / 4 = 495 - Then 792 and 924. At this point, you have: x^12 + 12x^11y + 66x^10y^2 + 220x^9y^3 + 495x^8y^4 + 792x^7y^5 + 924x^6y^6. From here, the coefficients mirror, so the final expansion is: x^12 + 12x^11y + 66x^10y^2 + 220x^9y^3 + 495x^8y^4 + 792x^7y^5 + 924x^6y^6 + 792x^5y^7 + 495x^4y^8 + 220x^3y^9 + 66x^2y^10 + 12xy^11 + y^12.
@jeveshjain4208
16 күн бұрын
Can’t you just use binomial theorem or even pascal’s triangle for this question?
@almasrifiras
16 күн бұрын
How?
@herbie_the_hillbillie_goat
16 күн бұрын
Yes
@oahuhawaii2141
13 күн бұрын
Yes, you can get the coefficients and powers for (x + y)¹² . But it's much easier to apply (x + y)³ first, then use (x' + y')² twice: 1:3:3:1 1:2:1 1:2:1 Cubing last is a big mess, so doing it early makes the numbers manageable. That is, I'd rather do (((√2-1)³)²)² instead of (17-12*√2)³ .
@oahuhawaii2141
13 күн бұрын
Yes, you can get the coefficients and powers for (x + y)¹² . But it's much easier to apply (x + y)³ first, then use (x' + y')² twice: 1:3:3:1 1:2:1 1:2:1 Cubing last is a big mess, so doing it early makes the numbers manageable. That is, I'd rather do (((√2-1)³)²)² instead of (17-12*√2)³ .
@oahuhawaii2141
13 күн бұрын
Yes, you can get the coefficients and powers for (x + y)¹² . But it's much easier to apply (x + y)³ first, then use (x' + y')² twice: 1:3:3:1 1:2:1 1:2:1 Cubing last is a big mess, so doing it early makes the numbers manageable. That is, I'd rather do (((√2-1)³)²)² instead of (17-12*√2)³ .
@MsBombastik
13 күн бұрын
just calculate it straight up. Yes when you know the answer and can notice all the correct simplifications(copying from someone else work) it becomes slightly shorter, but...
@timwood225
16 күн бұрын
An exemplary lesson in how to lay out a solution. Neat, organized, detailed, clear - the way it 'sposed to be. Which, when done, is itself a great aid to thinking.
@nonamenoname6921
14 күн бұрын
Although I’m not keen on his use of the phrase ‘cancels out’.
hmm but this is also a power rule situation is it not? I mean (sqrt(2)-1)^12= (((sqrt(2)-1)^2)^2))^3. In this look we can easily square the expression and reach a^2-2ab+b^2, which is 2 - 2*sqrt(2) - 1. Simplify to 1-2*sqrt(2). We square that again and then cube it. hungry to having issues following the formula, but i hope you get the gist of it. and then that should be a much simpler form.
@linsqopiring6816
15 күн бұрын
where you wrote "which is 2 - 2*sqrt(2) - 1" it should end in "+1" not "-1"
@kennethgee2004
15 күн бұрын
@@linsqopiring6816 thank you. i was hungry at the time posting so was messing up the math badly. The idea though was that you could use power rules evaluate this without getting crazily big numbers.
@linsqopiring6816
15 күн бұрын
@@kennethgee2004 No problem, and yea I think your way is less work. But I'm glad to see the approach taken in the video because it's an interesting system and it's good to know different ways to do something.
@oahuhawaii2141
13 күн бұрын
It's much easier to square for x², multiply for x³ = x*x², and square twice for x⁶ and then x¹² . The coefficients will be smaller when the "cubing" is done early; squaring isn't as bad. (((√2 - 1)³)²)² { cube it in my head } = ((5*√2 - 7)²)² { square it in my head } = (99 - 70*√2)² { must put down partial results } = 9801 + 9800 - (100-1)*140*√2 { manageable } = 19601 - 13860*√2
@Tarhaan
2 күн бұрын
One should use trigonometry and remember PI is useful
@RajMakwana-vg7hu
8 күн бұрын
Use Binomial Theorm
@xXspacecowboy2011Xx
4 күн бұрын
Not how I did it in my head. Use conjugate multiplication. X(srt 2+1)^12=2-1=1. Then x=1/((srt 2+1)^12).
@xXspacecowboy2011Xx
4 күн бұрын
(A+b)(a-b)= a^2-b^2.
@love-georgendon8071
4 күн бұрын
It's easier to use the Binomial Combination Expansion
@laitinlok1
9 күн бұрын
You can use binomial expansion. It's really easy anf the square identity is just the special cass.
@jtownball
Күн бұрын
Why not use the binomial expansion? Seems like that would have been just about as quick.
@louisdoumerc2693
14 күн бұрын
I did not used substitution for this test. Instead i did this (sqrt(2)-1)^12=((sqrt(2)-1)²(sqrt(2)-1)²)^3 after developing this expression i found the right solution.
@oahuhawaii2141
12 күн бұрын
It's easier to cube first, then square twice: (((√2 - 1)³)²)² If you don't remember how to cube, just square and then multiply: (((√2 - 1)²*(√2 - 1))²)² This is less tedious and less error-prone.
@paulortega5317
12 күн бұрын
Interesting is the relationship of this series of numbers to diophantine equations x² = 2y² + 1 and x² = 2y² - 1
@rvqx
16 күн бұрын
Do you think that is easier than just calculate it? (2^.5-1)²=3 - 2V2 (2^.5-1)⁴=17 - 12V2 (2^.5-1)⁸=577 - 408V2 (2^.5-1)**12=9809 - (17x408+12x577)V2 + 12x408x2=19601 - 13860V2
@Vega1447
16 күн бұрын
Egg zackly.
@mastnejbucek3411
16 күн бұрын
I agree with you but you forgot to calculate (add) 17x577 in the last line ;-)
@linsqopiring6816
15 күн бұрын
Congratulations, your application for Harvard has been approved!
@rvqx
15 күн бұрын
@@linsqopiring6816 Thank you. I am 76 , so i will go right away.
@rvqx
15 күн бұрын
@@mastnejbucek3411 17x577=9809 and adding 12x408x2 makes 19601
@tavaritsch
6 күн бұрын
Why should that equation be a problem? I just take out my calculator and put in the numbers for the result.🤔
@spicymickfool
16 күн бұрын
The term in parentheses is the solution to a quadratic equation. That implies the square of x is a constant times x plus another constant. But x^3 is the same first constant squared plus the constant term times x. With convenient expressions for x^2 and x^3 in only linear and constant terms, a 3 step iterative process gives the solution. I'd not think this is beyond the skill of A student in an introductory algebra course. Of course the binomial theorem would do it as well. I'm skeptical this comes from a Harvard interview.
@oahuhawaii2141
13 күн бұрын
Cube first, then square twice.
@xl000
11 күн бұрын
Stephen Hawking could do it at any power less than 4000.
@RobertoStenger
9 күн бұрын
I liked the approach very much. He gets to the heart of the problem and doesn't just want to solve it quickly.
@mrinaldas9614
14 күн бұрын
I did not take much time to calculate as x^6=(×^2×x^1)^2 X^12= (×^6)^2 Except for the last step,Idid not have to handle large nos
@oahuhawaii2141
13 күн бұрын
Yes, cube first, then square twice.
@ziyadullaabdiyev1971
14 күн бұрын
Assalomu alaykum. Rahmat sizga qiziqarli matematika uchun. Salomat bo‘ling
@Annoxy
9 күн бұрын
Binomial Theorem Left The Chat 💀💀
@RyanLewis-Johnson-wq6xs
14 күн бұрын
It’s in my head.
@anthonylau8821
4 күн бұрын
I have a strong feeling that if you do this in an uni interview, you will fail. University applicants can easier do the same by x^2 * x^2 * x^3 manually in 2 minutes. I wonder why the Harvard interviewer don't try x^96. Your method than will be significantly effective.
@xxnotmuchxx
13 күн бұрын
Hawkins looks high. Maybe thats where they got this channel name from.
@charlottesphie7037
2 күн бұрын
A simple mistake and you're done.
@valentinapaxley4221
8 күн бұрын
Oh my god. This is probably the only YT math I knew how to solve xD
@HollywoodF1
13 күн бұрын
I had to check- 19,601 and 13,860 have no common factors.
@oahuhawaii2141
13 күн бұрын
But the ratio is close to √2 .
@Teraverse420
16 күн бұрын
Bro forgot binomial theorem 😅😅😅
@stevereade4858
6 күн бұрын
Really? We know that the sqrt of 2 is 1.414 - 1 is 0.414 exp 12. We know it will be very small. Why waste all that time and mental energy?
@Gnowop3
13 күн бұрын
Quicker to calculate the square, then the 4th power then the 12th power. By the time you are half on your theory, I have already got the answer.
@oahuhawaii2141
13 күн бұрын
Your method of square, square, cube isn't as easy as cube, square, square. Try them out.
@Gnowop3
12 күн бұрын
@@oahuhawaii2141 not much different in terms of complexity if you know binomial expansion.
@99thminer
15 күн бұрын
Much simpler: (√2-1)^12 = x multiply both sides by (√2+1)^12 (2-1)^12 = 1 = x * (√2+1)^12 hence: x = 1 / (√2+1)^12
@vrijrajsaggu6245
8 күн бұрын
Just break the power to cube to square to square
@mandolinic
16 күн бұрын
I was going to do a calculator check, but you beat me to it! What a great question and a cool way of solving it.
@drwiz1968
16 күн бұрын
The presented solution is not the way to find the answer to the stated problem. Simply use the Binomial Theorem. As such you can do this on your iPhone, computer, etc. The days of drudgery are LONG over. In the future, make use of some clear thinking and readily available mathematical tools to render a solution.
@oahuhawaii2141
13 күн бұрын
Yes, please demonstrate this for the 12th power. If you can do error-free work for (x + y)¹² , then go for it! Most folks will make an error or more in the process, even given the coefficients and powers for (x + y)¹² . It's much easier to apply (x + y)³ first, then use (x' + y')² twice: 1:3:3:1 , 1:2:1 , 1:2:1 . Cubing last is a big mess, but doing it first keeps the numbers manageable. That is, I'd rather do (((√2-1)³)²)² instead of (17-12*√2)³ .
@mirkorokyta9694
16 күн бұрын
This is not higher mathematics ať all.
@williamwalker8107
5 күн бұрын
Nice algebraic reduction.
@socalvideo1
14 сағат бұрын
You're talking so fast I can't understand a thing you're saying.
@SALogics
14 күн бұрын
Tricky solution is here ✍
@user-wl2sp5rd3x
4 күн бұрын
Тонна никому ненужной "воды".
@aspenrebel
7 күн бұрын
2+2=5
@marceloboda4218
13 күн бұрын
Thank you guy
@llnovais
10 күн бұрын
Qual é o motivo de não ser: (√2 - 1)¹² = 2⁶ - 1 = 63
@SilentAdventurer
15 күн бұрын
Which Harvard interview?
@peterja6441
16 күн бұрын
now do it in reverse
@kateknowles8055
7 күн бұрын
🙃🙂😅😂🥴😄😄😀😄 Do the twelfth root of two instead; that is more help with musical scales.
@AjitSharma-km6ev
2 күн бұрын
Absolutely boring problem!
@JipsonMeitei
15 күн бұрын
Binomial
@oahuhawaii2141
13 күн бұрын
Please demonstrate this for the 12th power. If you can do error-free work for (x + y)¹² , then go for it! Most folks will make an error or 2 in the process, even given the coefficients and powers for (x + y)¹² . It's much easier to apply (x + y)³ first, then use (x' + y')² twice: 1:3:3:1 , 1:2:1 , 1:2:1 . Cubing last is a big mess, but doing it first makes the numbers manageable. That is, I'd rather do (((√2-1)³)²)² instead of (17-12*√2)³ .
@NoferTrunions
16 күн бұрын
A discussion of Significant Digits?
@oahuhawaii2141
13 күн бұрын
Yes, 19601 is close to 13860*√2 , and some calculators will not handle the math well.
@NoferTrunions
12 күн бұрын
@@oahuhawaii2141 Chem was all about sig figs. But when solving quadratic, the discriminant posed problems. My personal conclusion was in the aforementioned case to carry full precision and then apply sig figs at the end. The number 2 has 1 sig fig but its square root has infinite being irrational.
@angelmatematico45
16 күн бұрын
It's excelent but how do you formulate this problem in an exam, ao people do not apply diatributive
@rizwankhan-st5sl
13 күн бұрын
Use binomial theorem instead 😂
@oahuhawaii2141
13 күн бұрын
Please demonstrate this for the 12th power. If you can do error-free work for (x + y)¹² , then go for it! Most folks will make an error or 2 in the process, even given the coefficients and powers for (x + y)¹² . It's much easier to apply (x + y)³ first, then use (x' + y')² twice: 1:3:3:1 , 1:2:1 , 1:2:1 . Cubing last is a big mess, but doing it first keeps the numbers manageable. That is, I'd rather do (((√2-1)³)²)² instead of (17-12*√2)³ .
@ilafya
12 күн бұрын
Are you ok?
@meftahfraj8804
4 күн бұрын
Logx12 égal 12logx....
@dariusoberholster4246
16 күн бұрын
What if you just do Binomal Theorem over and over 12 times to get the answer?
@oahuhawaii2141
13 күн бұрын
Just cube first, then square twice. Try it out.
@aspenrebel
7 күн бұрын
Are u freakin' serious?
@olollol3346
9 күн бұрын
Isn't that dumb...
@abraham1376
16 күн бұрын
De la forma clasica es mas rapida
@jickey6108
16 күн бұрын
How do you take 18 minutes doing this
@davidbrisbane7206
16 күн бұрын
@@jickey6108 Because it takes a long time not to give an answer.
@linsqopiring6816
15 күн бұрын
check out the channel "tablet class math". Now that guy can talk. 20 min videos for much simpler than this.
@kateknowles8055
7 күн бұрын
It gives us time to be here reading and writing our comments, though.
@davidbrisbane7206
7 күн бұрын
@@kateknowles8055 Indeed 👍
@maximus0610
13 күн бұрын
In korea soso
@user-pp3ld4kb9s
16 күн бұрын
Nice. Where are you from. What is your name.
@cyirvine6300
16 күн бұрын
PLEASE!!! SLOW DOWN YOUR WORDS!!! I can't understand anything you're saying. I was hoping to brush up my math here, but it's pointless to listen to you. PAUSE BETWEEN STEPS AND SLOW DOWN.
@oahuhawaii2141
13 күн бұрын
You can hit the pause button and rewind and play again. There's also the settings to change the playback speed.
@CriticSimon
10 күн бұрын
Ahaha! Harvard? Sure
@hellogoodbye637
16 күн бұрын
Wtf didn't you solve for x and save yourself 9 mins of stupid calculations
@oahuhawaii2141
13 күн бұрын
The problem is to evaluate an expression, and he used x to refer to a particular value. There was no problem to solve for x.
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