A tricky problem from Harvard University Interview 

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.

@kennethgee2004: But a good engineer wouldn't say that because 2⁻¹² is an order of magnitude off from (√2 - 1)¹² .

@@oahuhawaii2141 🤣😂🤣🤣🤣😂👍

Ответ любителя: (√2-1)¹² более красиво и кратко выглядит, чем 19601-13860√2. 😊

And it's true

Im starting to relearn maths,.can you explain more in detail how the 12 power was fractioned in the second step

@@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

I like ur first way better

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

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).

so agreed

Hard math is.

@@aspenrebel I would love to see abstract algebra on this channel

@@xl000 pi r squared. No! Pie are round.

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

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)

@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

I was thinking the same just break the power into 2.2.3 and apply formula I got the same ans ❤👍

Better to cube, then square twice. And if the cube isn't easy, then square and then multiply.

It works and is faster

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

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...

Cool! Somebody tell Pythagoras.

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.

Very, very excellent solution procedure.

I have no idea what you did there.

It seems that your answer would give an negative solution? Not possible in x¹².

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

@@user-nr6tt5he7g ... Edited, thanks!

Квадрат бесконечности это бесконечность

Although I’m not keen on his use of the phrase ‘cancels out’.

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

How?

Yes

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 .

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 .

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 .

Yes, demonstrate this for the 12th power: (x + y)¹² = ???

@@oahuhawaii2141 Sure, I'd be happy to explain! Bear with me, since RU-vid 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.

(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.

@@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.

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.

where you wrote "which is 2 - 2*sqrt(2) - 1" it should end in "+1" not "-1"

@@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.

@@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.

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

(A+b)(a-b)= a^2-b^2.

Cube first, then square twice.

Yes, cube first, then square twice.

Egg zackly.

I agree with you but you forgot to calculate (add) 17x577 in the last line ;-)

Congratulations, your application for Harvard has been approved!

@@linsqopiring6816 Thank you. I am 76 , so i will go right away.

@@mastnejbucek3411 17x577=9809 and adding 12x408x2 makes 19601

Your method of square, square, cube isn't as easy as cube, square, square. Try them out.

@@oahuhawaii2141 not much different in terms of complexity if you know binomial expansion.

But the ratio is close to √2 .

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 .

Just cube first, then square twice. Try it out.

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 .

🙃🙂😅😂🥴😄😄😀😄 Do the twelfth root of two instead; that is more help with musical scales.

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 .

@@jickey6108 Because it takes a long time not to give an answer.

check out the channel "tablet class math". Now that guy can talk. 20 min videos for much simpler than this.

It gives us time to be here reading and writing our comments, though.

@@kateknowles8055 Indeed 👍

Yes, 19601 is close to 13860*√2 , and some calculators will not handle the math well.

@@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.

You can hit the pause button and rewind and play again. There's also the settings to change the playback speed.

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.

His x too😂