I did it using ((√2-1)^10) = (((√2-1)^2)^5) = (3 - 2√2)^5, the pascal triangle for i = 5 is 1 5 10 10 5 1, these are the coefficients for (a+b)^5 = a^5 + 5a^4b + 10 a^3 b^2 + 10 a^2 b^3 + 5 a b^4 + b^5 where a = 3 and b = -2√2, thus (a+b)^5 = 3363 - 2378√2.
I’m an 81 year old retired software developer and my cognition is not as good as it used to be. I really enjoyed watching this video and I thought you did a great job of explaining it. I then entered the information in my calculator and came up with the exact answer that you came up with and I thought wow.Keep up the good work.
Oh trust me, whether tour cognition is great or no, you shouldn't follow his terrible solution to find the answer. It's horribly slow and inefficient. Typical of those math videos involving exponents. Though, there's never a bad time to hone your skills. Hope you're doing okay in life besides that, coming from a youngin
I really enjoyed this video! But in this case, since the square root turns into a whole number when squared, I think it is actually easier just to power through using the original numbers, rather than defining 𝑥 = √2 - 1 and then substituting it back in the end: x¹⁰ = x⁸ ⋅ x² = [(x²)²]² ⋅ x² So, (√2 − 1)¹⁰ = {[(√2 − 1)²]²}² ⋅ (√2 − 1)² STEP 1: (√2 − 1)² = 3 − 2√2 STEP 2: (3 − 2√2)² = 17 − 12√2 STEP 3: (17 − 12√2)² = 577 − 408√2 STEP 4: (577 − 408√2) ⋅ (3 − 2√2) = 3363 − 2378√2
Since it's easier to work with low powers and do a final square, it would make sense to get the 5th power and then square that. The 5th power can be computed as ((x²)*x)*x² or ((x²)²)*x or x⁵ using the binomial expansion. 1st version: (√2-1)² = 3-2*√2 (√2-1)³ = (3-2*√2)*(√2-1) = 5*√2-7 (√2-1)⁵ = (3-2*√2)*(5*√2-7) = 29*√2-41 2nd version: (√2-1)² = 3-2*√2 (√2-1)⁴ = (3-2*√2)² = 17-12*√2 (√2-1)⁵ = (17-12*√2)*(√2-1) = 29*√2-41 3rd version: (√2-1)⁵ = 4*√2-5*4+10*2*√2-10*2+5*√2-1 = 29*√2-41 Final square: (√2-1)¹⁰ = ((√2-1)⁵)² = (29*√2-41)² = 841*2+1681 - 2*(1225-36)*√2 = 3363 - 2378*√2
Yeah man , i dont understand why they make simpler problems look like its super tough. Here wlso 2nd term of binomial is 1, so the expansion becomes more easily right?
@@sidharthpatra4751 I took it as [-(1-√2)]^10 so I got 1 as the first term, I don't have much experience with binomial equations, but I think that's faster?
@sidharthpatra4751 it's all about using this concept. If you used the binomial theorem for the 100th power it would get unmanageably tedious. This method would also be tedious, but a hell of a lot better.
Thanks for your clearly explained solution. I would approximate the answer in my head too by using (1.414 -1) = 0.414 which, raised to the tenth power, is almost zero, as a double-check that I had not made a mistake! Having a rough idea of what to expect is always a good check. (I've the advantage of reading so many comments with other methods, that I'm going to try some of these too.)
Let a be (SQRT(2)-1) then the equation can be factored into a² *( a²)² * (a²)²= a² * (a²)²)² a² is easy from 2nd binomial formula, squaring this a 2nd and 3rd time is not much more difficult. Now multiply 1st and 3rd term and you're finished.
Squaring is easy. Also, it's much easier to work with low powers first, and then do a final square. Thus, it would make sense to get the 5th power and then square that. The 5th power can be computed as ((x²)*x)*x² or ((x²)²)*x or x⁵ using the binomial expansion. 1st version: (√2-1)² = 3-2*√2 (√2-1)³ = (3-2*√2)*(√2-1) = 5*√2-7 (√2-1)⁵ = (5*√2-7)*(3-2*√2) = 29*√2-41 2nd version: (√2-1)² = 3-2*√2 (√2-1)⁴ = (3-2*√2)² = 17-12*√2 (√2-1)⁵ = (17-12*√2)*(√2-1) = 29*√2-41 3rd version: (√2-1)⁵ = 4*√2-5*4+10*2*√2-10*2+5*√2-1 = 29*√2-41 Final square: (√2-1)¹⁰ = ((√2-1)⁵)² = (29*√2-41)² = 841*2+1681 - 2*(1225-36)*√2 = 3363 - 2378*√2
I found a nice approach where you only need additions and substractions to get the exact formula. It's similar to the Fibonacci number algorithm. It's obvious that each whole power of (sqrt(2)-1) (refered to later as x) has the form a×sqrt(2)+b , hence we can encode them as a vector (a,b). We define an algorithm to compute all powers of x in a recursive way: a_n+1=b_n - a_n. b_n+1=a_n - a_n+1 That's because to get the next power, i.e. next element in the sequence we multiply by x: (a×sqrt(2)+b)×(sqrt(2)-1) = (b-a)sqrt(2) + (2a-b) = (b-a)sqrt(2) + (a -(b-a)) Note: We use a_n+1 for computing b_n+1 to avoid the multiplication by 2. The resulting sequence (as (a,b) vectors) looks like: (1,-1),(-2,3),(5,-7),(-12,17),(29,-41),(-70,99),(169,-239),(-408,577),(985,-1393),(-2378,3363) According to the 10th element of the sequence the 10th power is: -2378×sqrt(2) + 3363 IMO this approach is much more interesting than the video 😉
Very nice. In some sense no method can be easy because the answer is complicated. The recursion pattern can also be described well using 2 by 2 matrices. This sequence of numbers was collaterally known to Pythagoras and others because the ratios give good approximations to the square root of 2. The modern approach to all square roots is a subject in number theory that you can read about in Wikipedia under Pell’s Equation. The ratios are called continued fractions. The case of the Fibonacci numbers gives the best rational approximations to the golden ratio 😊
You needed a calculator anyway, so instead of “simplifying” over 17 minutes and the using a calculator, you could have used a calculator from the beginning and arrived to same result in less than 15 seconds ;) But I do love the arithmetics - always good to refresh :)
Which would have been marked wrong. The question was seeking the exact answer, not a calculator approximation. I know he fiddled about with a calculator at the end, but the problem was solved before that.
@@petermaling943 Exactly. The question was to simplify the equation, not to introduce inaccuracies ^^. Otherwise I can simplify PI to the number "3". Makes a lot things easier, but not right. That's the main difference between actual math and applied practical math. A carpenter who needs to calculate how long something needs to be just needs an appoximation that is close enough / has an acceptable margin of error. However that's not what math is about. The simplified equation can be evaluated / approximated when necessary to an arbitrary precision. The simplified forumla is the most accurate expression of the result.
@@petermaling943 My calculator was more accurate. 1.4867677997394958421382933796916e-4 That is so accurate that it would even be more accurate than would be necessary for quantum mechanics.
You can also write (sqrt(2)-1)=(1/x)*(1/(x+1/x)) with x=2^(1/4). Then, you use recursive expressions for (x+1/x)^n. That would make the result very general.
Amazing, how a simple calculation can be made this complicated! The square root of 2 is 1.4something. Subtract 1 and you get 0.4something. And then you multiply it with it self 10 times. Absolutely no need to introduce any variables!
Nice! Not sure if I quite made 2 minutes; probably more like 3 minutes here. I calculated the coefficients as 10 choose n instead of drawing up Pascal's triangle, as I thought it might be a tad faster; but I must be slowing in my dotage.
@@pietergeerkens6324 You just need a quarter of them for a good approximation: 1 10*9/2! = 45 10*9*8*7/4! = 720/24 * 7 = 210 -- 10*9*8*7*6*5/6! = (10 \over 4) = 210 (10 \over 8) = (10 \over 2) = 45 (10 \over 10) = (10 \over 0) = 1 If we consider 1 as given, you just have to calculate 45 and 210, which are pretty easy! Now the calculation/approximation: As I wrote above, with x = sqrt(2) - 1, 1/x = sqrt(2) + 1. So what is (1/x)^10? a = 1, b = 1 * sqrt(2) => a^n = 1, b^(2n) = 2^n makes it easy to calculate (a + b)^10: 1 * 2^5 + 45 * 2^4 + 210 * 2^3 + 210 * 2^2 + 45 * 2 + 1 = 32 + 720 + 1680 + 840 + 90 + 1 = 3363 (you can even make this simpler, if you calculate (a+b)^5 and square it, but this is easy enough to calculate!) So what's the approximation? x^10 = 1 / (1/x)^10 ~= 1 / (2 * 3363) = 1 / 6726 ~= 0.00014867678. Why 2 * 3363? Because (1 + sqrt(2))^10 = 3363 + c * sqrt(2). We haven't calculated c (spared us a lot of calculation!), but we know, that c * sqrt(2) ~= 3363. That's why we have (1 + sqrt(2))^10 ~= 2 * 3363 = 6726. The result is correct to the 11th digit or so! With some more calculation, we could even get the correct value, but why? While the text is very long now, it's not so much to do. Here's the kernel of the calculation: 10*9/2! = 45 10*9*8*7/4! = 210 (32 + 1) + (16 + 2) * 45 + (8 + 4) * 210 = 33 + 810 + 2520 = 3363 => x^10 ~= 1 / 6726 ~= 0.00014867678. Wow, that's even shorter than I thought! Which exponent hasn't he done yet? Would be as easy! E.g. (8 + 1) + (4 + 2) * 15 = 9 + 90 = 99 => x^6 ~= 1 / 198 ~= 0.0050505. (this was another video of this kind!)
Express in terms of binomials squared and evaluate each one. (r2-1)^10 =[ (r2-1)^2 ]^5 = (3-2r2)^5 but (3-2r2)^5 = [(3-2r2)^2]^2 (3-2r2) = (17-12r2)^2 (3-2r2 ) but (17-12r2)^2 (3-2r2) = (577-408r2) (3-2r2) and (577-408r2) (3-2r2) = 2547 - 2378r2 Hard to type!
I used the binomial expansion for the 5th power and then squared the result: (√2 - 1)¹⁰ = ((√2 - 1)⁵)² = (4*√2 - 5*4 + 10*2*√2 - 10*2 + 5*√2 - 1)² = (29*√2 - 41)² = 841*2+1681 - 2*(1225-36)*√2 = 3363 - 2378*√2
Let a=√2-1 and b=√2+1. ab=1; a+b=2√2. Using ab=1 without explicitly writing in the expansions below. (a+b)^2=a^2+b^2+2. a^2+b^2=6 (a^2+b^2)^2=a^4+b^4+2. a^4+b^4=34 (a^2+b^2)(a^4+b^4) = a^6+b^6+(a^2+b^2). a^6+b^6=6*33 (a^4+b^4)^2=a^8+b^8+2. a^8+b^8=34^2-2. (a^2+b^2)(a^8+b^8)=a^10+b^10+(a^6+b^6) a^10+b^10 = 6*(34^2-2)- 6*33 = 6*(34*33-1) = 6726. Note that a
@@brian554xx I saw the same thing as you and immediately recognized 6726 as twice 3363 (from the exact solution). I therefore had the same thought as you, and went straight to the comments to find someone asking your question or someone simply showing the method your 'saviour' (FatihKarakurt) showed. So, thanks, both of you!
Stumbled on this channel. Really enjoy your videos! The way you make a non-mathematician able to follow all steps is great. Your enthusiasm is fantastic - you just have to follow all steps through to see the final answer.
But the final answer still takes 6 computer operations to calculate a real number, just like the original. So is it really simplified or just a different format?
Well, for the expression (v-1)^10 = 3363 - 2378*v, the former requires one subtraction and then exponentiation which takes up log(n) multiplications. The latter is one multiplication and one subtraction. It's clearly a much easier expression to compute.
@@AvihooI Actually 3363-2378*v^0.5; It takes the same amount of time to do e^(0.5 * ln v) as for e^(10 * ln v); But if we did it all in assembly, I guess the simplified version is actually faster.
@@JaimeWarlock yeah I mean, if you formulate this problem using a logarithm and exponentiation which can be approximated using a Taylor expansion then indeed the simplifcation becomes redundant. This is why it depends on whatever computation model/system you're dealing with and what exactly is the problem you're trying to simplify. If the only tools you have are addition and multiplication (and subtraction and division) then the simplified form is better computationally. Btw I just noticed that in what I wrote, I forgot to mention that v = sqrt(2)... that expansion is obviously only true when that is the case.
I don't get the advantage of defining x this way, as you are transforming and transforming and transforming. I simply multiplied it out without defining such an x and I didn't need more lines than you had - I needed much less.
The answer may not be the goal, if it were we could use a calculators, the idea is to develop conceptual understanding. The long method is conceptual, it is pedagogically rich in delivering concepts.
@@moebadderman227 Then he isn't an engineer. Mathematics is the language of Engineering. We don't do four terms of calculus and linear algebra just for fun. BTW, the answer by op is incorrect, an ENGINEER would say 0.1487E-03 or 148.7E-06, since we use the Engineering notation and we round to at least 3 decimals when solving a practical problem. Also, if an expression uses roots it needs a lot more decimals to be precise. But that answer is still wrong, because there is no practical problem involved, it's pure math so rounding is not permitted. So final answer 3363 - 2378√2.
√2 =1.4142 is one of those numbers you memorise. The calculator will say it is 1.4142135124, so you can just go 0.4142^10 using your calculator and you get 0.000149 to 6 decimal places, using the 10 decimal places number you get 0.000149, which is the same? But of course the answer is to make x^2=1-2x and go thru the computations!!
I find it hard why the question was posed, and why not just use a calculator, Then I thought that they were seeking a number with no exponents or square roots in the final answer. The final answer still contained the square root of 2, so I really don't know what the point of the exercise was. However, it did require the solver to use some nice algebra techniques.
Very clear but 18 minute runtime could have been reduced to perhaps 6 minutes without losing viewers. There was no need, for example, to repeatedly write the same term on the left side of the equation from one line to the next. Also, different terms could be combined in a single step rather than combining the terms (e.g., constants) one at a time.
S,o do you mean that to solve this one should just understand to make some weird trick substitution? Looks like quite solveable with just brute force, as some have proven. But that brings us to one of the problem with some exam questions. How would one know when to make trick substitution and whe ? By chance, or by experience. Maybe their teacher just happened to show similar example on a lesson. Instead of trick questions I'd concentrate on testing routine in increasing difficulty levels.
I can’t help but be concerned about what school has only 6% of students getting this correct. Also, this is a fun exercise, but I think a direct approach with Pascal’s Triangle yields an answer more quickly and with fewer steps.
Chefe raiz cuadrada de 2 = (1,4142......) es un numero irracional, asi que si le resto 1 el resultado sigue irracional. Al ser elevado a la potencia 10 la parte entera tendra 10 digitos y la decimal infinitos ...
This is interesting, since 0 < sqrt(2) - 1 < 1 So raising this to larger and larger powers will tend to give numbers closer to zero. If you then get this in the form = Int_1 - Int_2 * sqrt(2), then Int_1 / Int_2 is a very good rational approximation for sqrt(2). In general this will be a method to get good rational approximations of numbers. For example if you want good rational approximations for sqrt(5), You could simply take (sqrt(5) - 2)^8 say and get the Int_1 and Int_2 and the approximation would be Int_1 / Int_2
Yes, a good insight. This can lead to Pell's equation, b² = na² + 1, whose solutions can give arbitrarily close rational approximations b/a ≈ √n for any non-square n>0. 1 = b² - na² = (b-a√n)(b+a√n) b-a√n = 1/(b+a√n) As a & b get larger, b-a√n gets smaller . . . Fred
I thought some elegant solution was asked for, but there is no elegance in your solution. There are many different ways to group it and multiply it out. Any of these solve it step by step in a rather basic way of using simple multiplication.
In todays world, where everyone has a calculator on the phone,the most "simplified" expression is the question itself. The solution is more complicated than the question. In 10 more years only 1% will try to complicate a clean written number like this..
(√2-1)^10 (√2-1)^2*(√2-1)^2^2^2 Then I just did it the long hard way. (√2-1)^2=3-2√2 (3-2√2)^2=17-12√2 (17-12√2)^2=577-408√2 (577-408√2)×(3-2√2)= 3364-2378√2 ~1.487×10^-4
This problem is nothing but a "donkey-work" one, with a very big chance to make a stupid arithmetic mistake and get a totaly wrong answer. There is a single way for a student to check the validity of his result (if he's not allowed to use a calculator): Since (√2-1)^10 = (1/(√2+1))^10 and 1/(√2+1) < 1/2, then: (√2-1)^10 < (1/2)^10 Therefore, the answer must satisfy: 0 < Answer ≪ 1 otherwise, there is a calculation error.
I don't understand. The solution does not look shorter or simpler than the problem. What is the point here? For such problem, I'd rather leave it as is or use a calculator.
Evaluate : (√2 - 1)¹⁰. There are a number of ways of doing this I shall use the Binomial Theorem Firstly I am going to do some preparation In order to reduce repetition and tedium. Now, in the Binomial Expansion of (√2 - 1)¹⁰. There are 11 terms. Each term has the following structure (binomial coefficient) x (a power of - √2 x (a power of 1) The powers of 1 are each equal to 1 So each term is now (a binomial coefficient) x (a power of √2 ) Every term with an even power of 2 is rational these powers are (L₁) (-√2)¹⁰=2⁵=32, (-√2)⁸=2⁴=16, (-√2 )⁶=2 ³=8, (-√2 )⁴= (2)²=4, (-√2 )²= 2 Every term with an odd power of √2 is irrational and can be written as a multiple of - √2 Thus (-√2)⁹= (-√2)⁸⋅〈-√2 )=(2)⁴.-√2 =-16√2 the complete list of these, (L₂) - 16√2, - 8√2, - 4√2, - 2√2, - √2.. ... (L₂) 32,16,8,4,2,1................ ..... ...(L₁) The Binomial Coefficients for power 10 are 1,10. 45, 120, 210, 252, 210, 120,45,10, 1.......... (L₃). (These form the 10th row of Pascal'sTriangle, or direct calculations using combinations). Now we can calculate the value of each term by multiplying each corresponding values in lists L₁, L₂ and L₃. ( 1x32 + 45x16+210x8+210x4+45x2+1)- √2 (10x16+120x8+252x4+120×2+10x1 Giving (32+720+1680+840+90+1) - (160+960+1008+240+10)√2 =3363 - 2378√2
if you change that to ((√2 - 1)^5)^2 you need only the coefficients +1 -5 +10 -10 +5 -1 and max exponent 5. After that sum up to get (29√2 - 41)^2 which is the result
When a number less than 1 is multiplied by itself, it gets smaller the more that is done. 10 times is a lot. A real smart person would say the answer is slightly above ZERO. Saving a ton time and move on to do something meaningful!
It depends a lot on the precision needed. For astronomy, even the precision given in the video may not be enough. For construction engineering, it's probably good enough to say ,00015, and it may oftentimes be enough to say very very small.
Everything that’s less than one (1) on a this large of an exponential isn’t all that much, actually it’s pretty close to zero. So the story was nice but the result is quite disappointing.
I did it using ((√2-1)^10) = (((√2-1)^2)^5) = (3 - 2√2)^5, the pascal triangle for i = 5 is 1 5 10 10 5 1, these are the coefficients for (a+b)^5 = a^5 + 5a^4b + 10 a^3 b^2 + 10 a^2 b^3 + 5 a b^4 + b^5 where a = 3 and b = -2√2, thus (a+b)^5 = 3363 - 2378√2.
exactly ehat I also did and it took only 5 minutes and no fiddlish substitution with a lot of probable mistakes
also is mine, more elegant
I did it very similarly. I expanded (a+b)^5 = (√2-1)^5 and simplified, yielding 29√2 - 41. Squaring that result gives 3363 - 2378√2
Why different result when this video count x2 = 1 - 2x. And u count x2 = 3-2root2😮
That was a lot of work that was verified by calculator at the end anyway! I guess using a calculator to solve wasn't permitted.
(√2 - 1)² = 3 - 2√2
(√2 - 1)^4 = (3 - 2√2)² = 17 - 12√2
(√2 - 1)^8 = (17 - 12√2)² = 577 - 408√2
(√2 - 1)^10 = (√2 - 1)²(√2 - 1)^8 = (3 - 2√2)(577 - 408√2) = 3363 - 2378√2
This is so much easier indeed.
That's how I did it. Didn't take 17 minutes, and I didn't use a calculator (except to check it at the end)
Simple...how to make a simple problem complex foolishly and create a video is what it's all about.
Toujours tellement LOURD avec lui !!!!!!!
@@renebrienne1862 நீங்கள் என்ன சொன்னிர்கள்?
I’m an 81 year old retired software developer and my cognition is not as good as it used to be. I really enjoyed watching this video and I thought you did a great job of explaining it. I then entered the information in my calculator and came up with the exact answer that you came up with and I thought wow.Keep up the good work.
Oh trust me, whether tour cognition is great or no, you shouldn't follow his terrible solution to find the answer. It's horribly slow and inefficient. Typical of those math videos involving exponents.
Though, there's never a bad time to hone your skills. Hope you're doing okay in life besides that, coming from a youngin
I really enjoyed this video! But in this case, since the square root turns into a whole number when squared, I think it is actually easier just to power through using the original numbers, rather than defining 𝑥 = √2 - 1 and then substituting it back in the end:
x¹⁰ = x⁸ ⋅ x² = [(x²)²]² ⋅ x²
So, (√2 − 1)¹⁰ = {[(√2 − 1)²]²}² ⋅ (√2 − 1)²
STEP 1: (√2 − 1)² = 3 − 2√2
STEP 2: (3 − 2√2)² = 17 − 12√2
STEP 3: (17 − 12√2)² = 577 − 408√2
STEP 4: (577 − 408√2) ⋅ (3 − 2√2) = 3363 − 2378√2
Since it's easier to work with low powers and do a final square, it would make sense to get the 5th power and then square that. The 5th power can be computed as ((x²)*x)*x² or ((x²)²)*x or x⁵ using the binomial expansion.
1st version:
(√2-1)² = 3-2*√2
(√2-1)³ = (3-2*√2)*(√2-1) = 5*√2-7
(√2-1)⁵ = (3-2*√2)*(5*√2-7) = 29*√2-41
2nd version:
(√2-1)² = 3-2*√2
(√2-1)⁴ = (3-2*√2)² = 17-12*√2
(√2-1)⁵ = (17-12*√2)*(√2-1) = 29*√2-41
3rd version:
(√2-1)⁵ = 4*√2-5*4+10*2*√2-10*2+5*√2-1
= 29*√2-41
Final square:
(√2-1)¹⁰ = ((√2-1)⁵)² = (29*√2-41)²
= 841*2+1681 - 2*(1225-36)*√2
= 3363 - 2378*√2
Since 10 is pretty small(small enough to calculate) you could do it with binomial theorem
That's how I did it! I got the answer, but it took some time. I think the method shown in the video is faster.
Yeah man , i dont understand why they make simpler problems look like its super tough. Here wlso 2nd term of binomial is 1, so the expansion becomes more easily right?
@@sidharthpatra4751 I took it as [-(1-√2)]^10 so I got 1 as the first term, I don't have much experience with binomial equations, but I think that's faster?
@sidharthpatra4751 it's all about using this concept. If you used the binomial theorem for the 100th power it would get unmanageably tedious. This method would also be tedious, but a hell of a lot better.
@@dekippiesip Uh ignore this comment
Thanks for your clearly explained solution. I would approximate the answer in my head too by using (1.414 -1) = 0.414 which, raised to the tenth power, is almost zero, as a double-check that I had not made a mistake! Having a rough idea of what to expect is always a good check. (I've the advantage of reading so many comments with other methods, that I'm going to try some of these too.)
(4√2-1)^5 => 4√2 - 5*4 +10*2√2 - 10*2 + 5*√2 - 1 (using pascal's triangle 1, 5, 10, 10, 5, 1)
=> (4+20+5)√2 - (20+20+1) => 29√2 - 41
^2 => 3363 - 2378√2
So beautiful! Brings back the old days when I was a student. We were not supposed to use a calculator in exams 😅
Let a be (SQRT(2)-1) then the equation can be factored into a² *( a²)² * (a²)²= a² * (a²)²)²
a² is easy from 2nd binomial formula, squaring this a 2nd and 3rd time is not much more difficult. Now multiply 1st and 3rd term and you're finished.
This is probably what most of us started with.
Squaring is easy. Also, it's much easier to work with low powers first, and then do a final square. Thus, it would make sense to get the 5th power and then square that. The 5th power can be computed as ((x²)*x)*x² or ((x²)²)*x or x⁵ using the binomial expansion.
1st version:
(√2-1)² = 3-2*√2
(√2-1)³ = (3-2*√2)*(√2-1) = 5*√2-7
(√2-1)⁵ = (5*√2-7)*(3-2*√2) = 29*√2-41
2nd version:
(√2-1)² = 3-2*√2
(√2-1)⁴ = (3-2*√2)² = 17-12*√2
(√2-1)⁵ = (17-12*√2)*(√2-1) = 29*√2-41
3rd version:
(√2-1)⁵ = 4*√2-5*4+10*2*√2-10*2+5*√2-1
= 29*√2-41
Final square:
(√2-1)¹⁰ = ((√2-1)⁵)² = (29*√2-41)²
= 841*2+1681 - 2*(1225-36)*√2
= 3363 - 2378*√2
My math must suck. .4142^10 is a really small number and doesn't match your result at all. Oh well
I found a nice approach where you only need additions and substractions to get the exact formula. It's similar to the Fibonacci number algorithm.
It's obvious that each whole power of (sqrt(2)-1) (refered to later as x) has the form a×sqrt(2)+b , hence we can encode them as a vector (a,b).
We define an algorithm to compute all powers of x in a recursive way:
a_n+1=b_n - a_n.
b_n+1=a_n - a_n+1
That's because to get the next power, i.e. next element in the sequence we multiply by x: (a×sqrt(2)+b)×(sqrt(2)-1) = (b-a)sqrt(2) + (2a-b) = (b-a)sqrt(2) + (a -(b-a))
Note: We use a_n+1 for computing b_n+1 to avoid the multiplication by 2.
The resulting sequence (as (a,b) vectors) looks like:
(1,-1),(-2,3),(5,-7),(-12,17),(29,-41),(-70,99),(169,-239),(-408,577),(985,-1393),(-2378,3363)
According to the 10th element of the sequence the 10th power is: -2378×sqrt(2) + 3363
IMO this approach is much more interesting than the video 😉
This is really ingenious, congrat!
@@zk6912 Thank you
Very nice.
In some sense no method can be easy because the answer is complicated. The recursion pattern can also be described well using 2 by 2 matrices. This sequence of numbers was collaterally known to Pythagoras and others because the ratios give good approximations to the square root of 2. The modern approach to all square roots is a subject in number theory that you can read about in Wikipedia under Pell’s Equation. The ratios are called continued fractions. The case of the Fibonacci numbers gives the best rational approximations to the golden ratio 😊
Collaterally is a misprint/autocorrect. Just skip it.
It's just a simple method to track the coefficients as you successively multiply each power by (√2-1) to get the next power.
You needed a calculator anyway, so instead of “simplifying” over 17 minutes and the using a calculator, you could have used a calculator from the beginning and arrived to same result in less than 15 seconds ;)
But I do love the arithmetics - always good to refresh :)
Which would have been marked wrong. The question was seeking the exact answer, not a calculator approximation. I know he fiddled about with a calculator at the end, but the problem was solved before that.
@@petermaling943 Exactly. The question was to simplify the equation, not to introduce inaccuracies ^^. Otherwise I can simplify PI to the number "3". Makes a lot things easier, but not right. That's the main difference between actual math and applied practical math. A carpenter who needs to calculate how long something needs to be just needs an appoximation that is close enough / has an acceptable margin of error. However that's not what math is about. The simplified equation can be evaluated / approximated when necessary to an arbitrary precision.
The simplified forumla is the most accurate expression of the result.
@@petermaling943 My calculator was more accurate. 1.4867677997394958421382933796916e-4 That is so accurate that it would even be more accurate than would be necessary for quantum mechanics.
(2^1/2)^10 * (1-1/2^1/2)^10 Since 1/1/2^1/2 is less than 1 , use expansion series or (1-x)^n. First three terms * 32 gives you approx answer
Just use Pascal's triangle to get the binomial coefficients and simplify. Easy.
Expanding the Pascal triangle to the 10th power is the longest part. After that the result just drops out in about 4 lines of calculation.
You can also write (sqrt(2)-1)=(1/x)*(1/(x+1/x)) with x=2^(1/4). Then, you use recursive expressions for (x+1/x)^n. That would make the result very general.
To further … this is related to Lucas sequence (see Wikipedia)
Demonstrates the incredible accuracy of modern calculators. The difference of too very similar values will show any errors propagation.
Amazing, how a simple calculation can be made this complicated! The square root of 2 is 1.4something. Subtract 1 and you get 0.4something. And then you multiply it with it self 10 times. Absolutely no need to introduce any variables!
Multiply with itself 9 times. You also need more precision to make sure the result doesn't deviate from the true answer by too much.
I assume you're an engineer rather than a mathematician.
Square both terms inside the paratehecies and the power 10; You have (2-1)* or 1 to the 100th power which is 1 = 1.
Pascal triangle... 2 minutes, exact solution
Absolutely. No faffing around witth fancy methods that only work for a very restricted set of problems.
Nice! Not sure if I quite made 2 minutes; probably more like 3 minutes here. I calculated the coefficients as 10 choose n instead of drawing up Pascal's triangle, as I thought it might be a tad faster; but I must be slowing in my dotage.
Make a video :-)
I'd rather say 5-10 minutes.
Half, if you're clever.
@@pietergeerkens6324 You just need a quarter of them for a good approximation:
1
10*9/2! = 45
10*9*8*7/4! = 720/24 * 7 = 210
--
10*9*8*7*6*5/6! = (10 \over 4) = 210
(10 \over 8) = (10 \over 2) = 45
(10 \over 10) = (10 \over 0) = 1
If we consider 1 as given, you just have to calculate 45 and 210, which are pretty easy!
Now the calculation/approximation:
As I wrote above, with x = sqrt(2) - 1, 1/x = sqrt(2) + 1. So what is (1/x)^10?
a = 1, b = 1 * sqrt(2) => a^n = 1, b^(2n) = 2^n makes it easy to calculate (a + b)^10:
1 * 2^5 + 45 * 2^4 + 210 * 2^3 + 210 * 2^2 + 45 * 2 + 1 = 32 + 720 + 1680 + 840 + 90 + 1 = 3363
(you can even make this simpler, if you calculate (a+b)^5 and square it, but this is easy enough to calculate!)
So what's the approximation? x^10 = 1 / (1/x)^10 ~= 1 / (2 * 3363) = 1 / 6726 ~= 0.00014867678.
Why 2 * 3363? Because (1 + sqrt(2))^10 = 3363 + c * sqrt(2). We haven't calculated c (spared us a lot of calculation!), but we know, that c * sqrt(2) ~= 3363. That's why we have (1 + sqrt(2))^10 ~= 2 * 3363 = 6726.
The result is correct to the 11th digit or so!
With some more calculation, we could even get the correct value, but why?
While the text is very long now, it's not so much to do. Here's the kernel of the calculation:
10*9/2! = 45
10*9*8*7/4! = 210
(32 + 1) + (16 + 2) * 45 + (8 + 4) * 210 = 33 + 810 + 2520 = 3363 => x^10 ~= 1 / 6726 ~= 0.00014867678.
Wow, that's even shorter than I thought! Which exponent hasn't he done yet? Would be as easy!
E.g. (8 + 1) + (4 + 2) * 15 = 9 + 90 = 99 => x^6 ~= 1 / 198 ~= 0.0050505. (this was another video of this kind!)
I was just curious:
(n=2): (2 + 1) = 3 => x^2 ~= 1 / 6 ~= 0.166667. Exact value (this is the worst approximation!): 0.171573.
(n=14): 14*13/2! = 91, 14*13*12*11/4! = 91*11 = 1001, 14*13*12*11*10*9/6! = 1001*3 = 3003.
(128 + 1) + (64 + 2) * 91 + (32 + 4) * 1001 + (16 + 8) * 3003 = 129 + 6006 + 36036 + 72072 = 114243
=> x^14 ~= 1 / 228486 ~= 0.0000043766357676. Exact value: 0.0000043766357677. 16th digit!
Reciprocal value is 228485.9999956233642323, which is pretty close to 228486 (even more 228486 - x^14).
Express in terms of binomials squared and evaluate each one.
(r2-1)^10 =[ (r2-1)^2 ]^5 = (3-2r2)^5
but (3-2r2)^5 = [(3-2r2)^2]^2 (3-2r2) = (17-12r2)^2 (3-2r2 )
but (17-12r2)^2 (3-2r2) = (577-408r2) (3-2r2)
and (577-408r2) (3-2r2) = 2547 - 2378r2
Hard to type!
I used the binomial expansion for the 5th power and then squared the result:
(√2 - 1)¹⁰
= ((√2 - 1)⁵)²
= (4*√2 - 5*4 + 10*2*√2 - 10*2 + 5*√2 - 1)²
= (29*√2 - 41)²
= 841*2+1681 - 2*(1225-36)*√2
= 3363 - 2378*√2
The procedure shown ignores that we deal here with irrational numbers which have infinitely many decimals!
@@AljosaVolcic: And why is that significant?
exactly my approach as well. Found it much simpler and failure proof
(1.41-1)¹⁰
(.41)¹⁰= .00015
@praveencg5828: Your calculator is broken:
(1.41 - 1)¹⁰ = 0.41¹⁰
= 0.00013422659310152401
(√2 - 1)¹⁰ = 3363 - 2378*√2
≈ 0.0001486767799739495842138293...
Really close to 1/6726. What's the simplest way to get this answer algebraically?
Let a=√2-1 and b=√2+1. ab=1; a+b=2√2. Using ab=1 without explicitly writing in the expansions below.
(a+b)^2=a^2+b^2+2. a^2+b^2=6
(a^2+b^2)^2=a^4+b^4+2. a^4+b^4=34
(a^2+b^2)(a^4+b^4) = a^6+b^6+(a^2+b^2). a^6+b^6=6*33
(a^4+b^4)^2=a^8+b^8+2. a^8+b^8=34^2-2.
(a^2+b^2)(a^8+b^8)=a^10+b^10+(a^6+b^6)
a^10+b^10 = 6*(34^2-2)- 6*33 = 6*(34*33-1) = 6726.
Note that a
@@FatihKarakurt I love you. Please continue to exist!
@@brian554xx
I saw the same thing as you and immediately recognized 6726 as twice 3363 (from the exact solution). I therefore had the same thought as you, and went straight to the comments to find someone asking your question or someone simply showing the method your 'saviour' (FatihKarakurt) showed. So, thanks, both of you!
@@FatihKarakurt
I knew it! ✊️🎉
It had to be the conjugate at play! 🔍☝️🧐 ...😆😃
Thanks for doing the work! 🙏💯
Stumbled on this channel. Really enjoy your videos! The way you make a non-mathematician able to follow all steps is great. Your enthusiasm is fantastic - you just have to follow all steps through to see the final answer.
But the final answer still takes 6 computer operations to calculate a real number, just like the original. So is it really simplified or just a different format?
Well, for the expression (v-1)^10 = 3363 - 2378*v, the former requires one subtraction and then exponentiation which takes up log(n) multiplications. The latter is one multiplication and one subtraction. It's clearly a much easier expression to compute.
@@AvihooI Actually 3363-2378*v^0.5; It takes the same amount of time to do e^(0.5 * ln v) as for e^(10 * ln v); But if we did it all in assembly, I guess the simplified version is actually faster.
@@JaimeWarlock yeah I mean, if you formulate this problem using a logarithm and exponentiation which can be approximated using a Taylor expansion then indeed the simplifcation becomes redundant.
This is why it depends on whatever computation model/system you're dealing with and what exactly is the problem you're trying to simplify. If the only tools you have are addition and multiplication (and subtraction and division) then the simplified form is better computationally.
Btw I just noticed that in what I wrote, I forgot to mention that v = sqrt(2)... that expansion is obviously only true when that is the case.
I don't get the advantage of defining x this way, as you are transforming and transforming and transforming. I simply multiplied it out without defining such an x and I didn't need more lines than you had - I needed much less.
By rasing to higher powers, you can calculate V2 (approximately 3363/2378 here)
Thank you for the explanation of the equation. I am going to try this equation a few times until I can figure it out. I appreciate the equation
The answer may not be the goal, if it were we could use a calculators, the idea is to develop conceptual understanding. The long method is conceptual, it is pedagogically rich in delivering concepts.
Engineer says 1.5E-4
My brother is an Engineer, and I admire him and his fellow Engineers for making things I could never make; but Mathematics? No.
@@moebadderman227 Then he isn't an engineer. Mathematics is the language of Engineering. We don't do four terms of calculus and linear algebra just for fun. BTW, the answer by op is incorrect, an ENGINEER would say 0.1487E-03 or 148.7E-06, since we use the Engineering notation and we round to at least 3 decimals when solving a practical problem. Also, if an expression uses roots it needs a lot more decimals to be precise.
But that answer is still wrong, because there is no practical problem involved, it's pure math so rounding is not permitted. So final answer 3363 - 2378√2.
@@bigboss9817
"Then he isn't an engineer."
O RLY
A binomial theorem is the common solution to that expression, but your solution is so amazing. Thank you sir
could you use pascals triangle for this instead?
The starting form is simple enough, certainly more than the result and the process leading to it :)
That was super interesting. Thanks for taking the time to make this video. ❤
Shouldn’t the rounding off result in 0.0001487?
To many math?
I'd compute (sqrt(2)-1)^2 = 3-2sqrt(2).
And then raise to 5th power your way (4+1).
No x.
I'd raise it to the 5th power and then square the result.
√2 =1.4142 is one of those numbers you memorise. The calculator will say it is 1.4142135124, so you can just go 0.4142^10 using your calculator and you get 0.000149 to 6 decimal places, using the 10 decimal places number you get 0.000149, which is the same? But of course the answer is to make x^2=1-2x and go thru the computations!!
Shorter to develop and calculate directly. No?
How long willmit take with the binomial theorem?
I've done binomial expansion for this one... the other way is obviously a nice shortcut.
I find it hard why the question was posed, and why not just use a calculator, Then I thought that they were seeking a number with no exponents or square roots in the final answer. The final answer still contained the square root of 2, so I really don't know what the point of the exercise was. However, it did require the solver to use some nice algebra techniques.
Thanks man.
HP prime in CAS mode and Wolfram alpha gives the answer integer and squareroot answer
Very clear but 18 minute runtime could have been reduced to perhaps 6 minutes without losing viewers. There was no need, for example, to repeatedly write the same term on the left side of the equation from one line to the next. Also, different terms could be combined in a single step rather than combining the terms (e.g., constants) one at a time.
Pop quiz: is E actually equal to MC/2? I dont think so. C is the neasurement of the speed of something, which is disqualifying for this equation.
S,o do you mean that to solve this one should just understand to make some weird trick substitution? Looks like quite solveable with just brute force, as some have proven. But that brings us to one of the problem with some exam questions. How would one know when to make trick substitution and whe ? By chance, or by experience. Maybe their teacher just happened to show similar example on a lesson. Instead of trick questions I'd concentrate on testing routine in increasing difficulty levels.
Textbook binomial theorem question
What kind of pen are you using?
Profesor: al hacer la operación con 3363-2378 por la raíz cuadrada de 2 = 1.4867677997 . a ver si está correcta esta operación.
can b done by binomial expansion .
I enjoyed the video as much as the comments. Thank you to all of you !!
Since when is six squared equal to 25??
(sqrt2-1^10
= (2 - 2sqrt2 + 1)^5
= (3 - 2sqrt2)(9 - 12sqrt2 + 8)^2
= (3 -2sqrt2)(289 - 408sqrt2 + 288)
= (3 -2sqrt2)(577 - 408sqrt2)
= (1731+1632-1224sqrt2-1554sqrt2)
= (3363 - 2778sqrt2)
I don't get the point of all this... Was the objective to turn the original expression into one without exponents? Unclear
no point to complicate things with x
Since the numbers are not that complicated, I would probably. have tried taking the cube of the cube then multiply with sqt2-1
We still need a calculator to evaluate the final answer so why not use a calculator in the first place?
Nice to see the force power and the tense power
You people in the comments are so much smarter than I. I am trying really hard but I can’t grasp why this equation is difficult!
Line diagnal /2×4 (-n!)
I can’t help but be concerned about what school has only 6% of students getting this correct. Also, this is a fun exercise, but I think a direct approach with Pascal’s Triangle yields an answer more quickly and with fewer steps.
Hang on. Right at the end you used a calculator. So why all other stuff ?
I get the whole logic, but i dont get how you do some of those simplifications. I wish i knew what to look for to understand.
Chefe raiz cuadrada de 2 = (1,4142......) es un numero irracional, asi que si le resto 1 el resultado sigue irracional. Al ser elevado a la potencia 10 la parte entera tendra 10 digitos y la decimal infinitos ...
I could have expanded it with distribution by the time I worked through this convoluted method.
This is interesting, since 0 < sqrt(2) - 1 < 1
So raising this to larger and larger powers will tend to give numbers closer to zero.
If you then get this in the form = Int_1 - Int_2 * sqrt(2), then Int_1 / Int_2 is a very good rational approximation for sqrt(2). In general this will be a method to get good rational approximations of numbers.
For example if you want good rational approximations for sqrt(5),
You could simply take (sqrt(5) - 2)^8 say and get the Int_1 and Int_2 and the approximation would be Int_1 / Int_2
Yes, a good insight.
This can lead to Pell's equation, b² = na² + 1, whose solutions can give arbitrarily close rational approximations b/a ≈ √n for any non-square n>0.
1 = b² - na² = (b-a√n)(b+a√n)
b-a√n = 1/(b+a√n)
As a & b get larger, b-a√n gets smaller . . .
Fred
I thought some elegant solution was asked for, but there is no elegance in your solution. There are many different ways to group it and multiply it out. Any of these solve it step by step in a rather basic way of using simple multiplication.
good demonstration
x^4 = 17 - 12 sqrt(2)
x^6 = 99 - 70 sqrt(2)
x^10 gives the same result
By inspection: "A very small number" (values less than 0.5 to the power of 10 = very small indeed)
I solved it using Pascal's Triangle, fun problem
How was that simpler?
Binomial theorem with pascle triangle expansion will do the quick trick to solve …🧐
In todays world, where everyone has a calculator on the phone,the most "simplified" expression is the question itself. The solution is more complicated than the question. In 10 more years only 1% will try to complicate a clean written number like this..
I love when the question says "Simplify" and the final answer looks 100x more complicated than the initial equation LOL
Use Tartaglia triangle....it takes less
Binomial expansion much?
I used my HP11C and got the answer in five seconds.
(√2-1)^10
(√2-1)^2*(√2-1)^2^2^2
Then I just did it the long hard way.
(√2-1)^2=3-2√2
(3-2√2)^2=17-12√2
(17-12√2)^2=577-408√2
(577-408√2)×(3-2√2)=
3364-2378√2
~1.487×10^-4
This problem is nothing but a "donkey-work" one, with a very big chance to make a stupid arithmetic mistake and get a totaly wrong answer.
There is a single way for a student to check the validity of his result (if he's not allowed to use a calculator):
Since (√2-1)^10 = (1/(√2+1))^10 and 1/(√2+1) < 1/2, then: (√2-1)^10 < (1/2)^10
Therefore, the answer must satisfy: 0 < Answer ≪ 1 otherwise, there is a calculation error.
~ 82^-2
~ 6726^-1
I don't understand. The solution does not look shorter or simpler than the problem. What is the point here? For such problem, I'd rather leave it as is or use a calculator.
Love you the way you draw X
Why you are not using binomal formula, you will get result in 5 minutes.
Not sure whether the final answer is “simpler” than the original…
Evaluate : (√2 - 1)¹⁰.
There are a number of ways of doing
this
I shall use the Binomial Theorem
Firstly I am going to do some
preparation
In order to reduce repetition and
tedium.
Now, in the Binomial Expansion of
(√2 - 1)¹⁰.
There are 11 terms.
Each term has the following structure
(binomial coefficient) x (a power of - √2 x (a power of 1)
The powers of 1 are each equal to 1
So each term is now (a binomial coefficient) x (a power of √2 )
Every term with an even power of 2 is rational these powers are (L₁)
(-√2)¹⁰=2⁵=32, (-√2)⁸=2⁴=16, (-√2 )⁶=2 ³=8, (-√2 )⁴= (2)²=4, (-√2 )²= 2
Every term with an odd power of √2 is irrational
and can be written as a multiple of - √2
Thus (-√2)⁹= (-√2)⁸⋅〈-√2 )=(2)⁴.-√2 =-16√2 the complete list of these, (L₂)
- 16√2, - 8√2, - 4√2, - 2√2, - √2.. ... (L₂)
32,16,8,4,2,1................ ..... ...(L₁)
The Binomial Coefficients for power 10 are
1,10. 45, 120, 210, 252, 210, 120,45,10, 1.......... (L₃).
(These form the 10th row of Pascal'sTriangle, or direct calculations
using combinations).
Now we can calculate the value of each term by multiplying each
corresponding values in lists L₁, L₂ and L₃.
( 1x32 + 45x16+210x8+210x4+45x2+1)-
√2 (10x16+120x8+252x4+120×2+10x1
Giving (32+720+1680+840+90+1)
- (160+960+1008+240+10)√2
=3363 - 2378√2
if you change that to ((√2 - 1)^5)^2 you need only the coefficients +1 -5 +10 -10 +5 -1 and max exponent 5. After that sum up to get (29√2 - 41)^2 which is the result
When a number less than 1 is multiplied by itself, it gets smaller the more that is done.
10 times is a lot. A real smart person would say the answer is slightly above ZERO.
Saving a ton time and move on to do something meaningful!
It depends a lot on the precision needed. For astronomy, even the precision given in the video may not be enough. For construction engineering, it's probably good enough to say ,00015, and it may oftentimes be enough to say very very small.
You can divide expression by 2 and use trigonometry to solve it elegantly, since 1/sqrt2 and 1/2 are sines of angles.
Totally confusing. But then people really good at maths are generally pretty confused.
Interesting...... I like maths, but does it makes sense to replace one term by another one which looks optically worse?
as i am an engeener the answer come quicly ~0 😂 because √2 - 1 is around 0.414 that is
Everything that’s less than one (1) on a this large of an exponential isn’t all that much, actually it’s pretty close to zero. So the story was nice but the result is quite disappointing.
I'll take your word for it.
why not just take log square root 2-i by 10 and find anti log
X to the tense power...it was a very nervous power...sorry I couldn't resist.
Beautiful question! Thank you!
At 07:00, you said "168" but wrote "169".
Take log 10 log(1.414-1)
10 log 0.414
Students of what get this wrong? It was 25 years ago but i remember EE was more difficult than this...
Brother it could be solved simply by using the binomial theorem
Is this the best voice on youtube???
I'm confused. What does the upside down Y represent?