*_Don't forget: 10% of my Ad and Merch Income this month will go to humanitarian help for Ukraine. So please make sure to watch the video in its entirety and to check out some of today's relevant links! :)_* Train your Algebra Expertise by trying out Brilliant! =D brilliant.org/FlammableMaths Support the channel by checking out Deez Nutz over on stemerch.eu/ ! :3 Engi Watch: stemerch.com/products/the-incredibly-unrigorous-engineering-watch?variant=40377075728562 Wuck. play.google.com/store/apps/details?id=org.flammablemaths.Wuck Check out my newest video over on @Flammy's Wood ! =D th-cam.com/video/tTpjWePfK7o/w-d-xo.html Shirt from the Video: stemerch.eu/collections/fugacity
As a Ukrainian, I would ask to send money not for humanitarian aid, but for the army. The main donations go to humanitarian (several times more), although the only thing that can save cities with civilians is the army. For example, you can donate to the "turn back alive"(повернись живим) fund , or any other that sends money to the army or to hospitals.
I imagine that the intent was to demonstrate properties of exponentiation, because the initial equation is just the classic quadratic x^2-2x+1=0 in disguise, which only has solution = 1, so of course 1^m+1/(1^n)=2 for any powers m and n.
I love the method you use because you could easily solve for x and plug in the value. But in this way you get to the solution without needing to know what x is. It's mathemagical.
@@PapaFlammy69 Hello , i like your videos check this out because there are two ways to disprove it, one when x€R and x€C. Let this equation be; x²+x+1=0 (1) , x≠0 Then we divide (1) by x; x+1+1/x=0 From the first (1) equation we solve for x+1; x²+x+1=0=> x+1=-x² (2) Now we have; x+1 +1/x=0 =>(2) -x²+1/x=0=> x²=1/x Now we multiply by x (x≠0) and he have; x³=1=> x=1. But if x=1; (1)=> x²+x+1=0=> 1²+1+1=0=> 1+1+1=0=> 3=0. That's the "problem" but there are two ways to disprove it. I Found out one but the other when x€C was a bit confusing. I would appreciate if you answered or make a small video. (I am fan of your work). Have a nice day!
I think that is less interesting. Yeah, it solves the problem, but it is more beautiful to see the pattern that occurs and the mystery of solving the problem without the actual value of x.
I have managed to solve this for the general case. This will be a long one, so buckle up. So, let's say we have a(1) = x + x^-1 = k (we'll have a detailed discussion about k later) and we want to find the value of a(n) = x^n + x^-n. We're well aware that if n = 0, then a(0) = x^0 + x^-0 = 2, and for n = 1, it's given. Then, if we want to find a(n + 1), we first multiply (x + x^-1)(x^n + x^-n) = a(1)a(n) = x^(n + 1) + x^-(n+1) + x^(n - 1) + x^-(n - 1) = a(n + 1) + a(n - 1). We know a(1), it's k, so our recurrence relation becomes a(n + 1) = k*a(n) - a(n - 1). This is really important, because we can now use familiar linear recurrence solving techniques, because this is a linear recurrence with constant coefficients (k is constant w.r.t. this equation), a second order one at that (because of a(n) and a(n-1)). We can use the characteristic polynomial, it will be t^2 - kt + 1 = 0 which gives us t = k/2 ± √((k/2)² - 1). If we want t to be real, then (k/2)² - 1 >= 0 or (k/2)² >= 1, so k >= 2 or k
Many thanks for this: I'd got to the same formula for a(n) in a more brute-force way. In answer to your conjecture, I'm fairly sure that if k is an integer, then a(n) is an integer too - because the odd powers of the surds in the two parts of the formula cancel, leaving only the even powers. I'm now wondering whether there's a fairly simple closed formula for a(n) in terms of powers of k with integer coefficients.
If you let x equal some complex number e^iy, then x + 1/x = 2cos(y) and x^n + 1/x^n = 2cos(ny). Then you’re rephrasing the question into trying to find 2cos(ny) in terms of 2cos(y), which can be found by doing the real part of the expansion (cosx + isinx)^n. You might then be able to generalise
At first I use the same method as you did. But later on I found that using AM-GM inequality is much easier. x+1/x≥2√x(1/x) x+1/x≥2 From this,we know that x+1/x=2 when x=1.So x^5+1/x^5=2
The only problem is that, for AM-GM x>0 must be given. (you could consider a case where it is greater than 0 and less than 0 and then get a full condition as follows: 1. if x is greater than 0 then, x+1/x > 2 (or equal) and 2. if x is less than 0, x+1/x < -2 (or equal)
Papa Flammy I have an idea that might involve more than one person. I really love that incredibly unrigorious engineering watch, however I wouldnt buy it becuase I already have a smartwatch and I would switch that. Howwever, I think that you could make that(or even other maths watch faces) for smartwatches, people would probably pay for it. I know there are already some out there, but I think we could use more funny desings like seen on the engineering watch
It just struck me: you could use am-gm inequality to solve this. However you'll only get a solution in the set of positive real numbers...... x+1/x ≥ 2 that means that x= 1/x = 1 Then x^5 + 1/x^5 = 2 Edit: the answer cannot have a negative value because if x < 0 then 1/x < 0 and hence x + 1/x < 0
Take the original equation, multiply all sides by x, switch it around, and we have x^2-2x+1=0, or (x-1)^2=0. So then x=1 and we can substitute this into the x^5 equation to get 2
Actually only if both the numbers are natural numbers if integers are consider 91 is the smallest number which can be expressed as sum of two cubes in two ways that is : 3³+4³=6³+(-5)³ .
This is boundary value problem , minimum positive value of x + 1/x is 2 which can be proved by am gm inequality so the value of x is 1 then put in x^5 + 1/x^5 which will give 1+ 1 = 2
I received a very similar question to this from SMO senior a little while before I saw this vid. I was able to solve it using your technique. thanks papa!
Hi, if interested in math competitions (including SMO)? If so, take a look at th-cam.com/video/rkzxdMFEEtw/w-d-xo.html and others in the Olympiad playlist. You will see a lot of tricky and somewhat complex problems to try. Hope to see you there.
I am not well versed in the formalism of mathematics and idk if the following is valid. Generalize this problem to the complex world, hence z+1/z=2 if we limit ourselves to the unit circle in the complex plane, we get r = 1. [idk if we can do this or not without losing the fact that x is not bounded] 2cos(a)=2 -> a=2*pi*k where k∈Z for the 5th case z^5 + z^(-5) 2cos(5a) = 2 hence x^5 + 1/x^5 = 2 ? again, I have no idea about formalisms and errors of assuming it to be in a restricted part of the complex plane. But it works?
we can just use cauchy inequality for this also! x + 1/x >= 2√x.1/x = 2, and we also hav x + 1/x = 2 => x = 1/x => x = 1 so we can lead that to the final answer !
The graph for x+(1/x) is very cool check it out Here in india we are taught how to find the range of several functions And this is one of the infamous ones It's range goes as (-infinity,-2] U [2,Inf) Where -2 and 2 occur and -1 and 1 respectively
I personally just solved for x and subbed in, but your way of solution really got me thinking. Guess it is the difference between an A student and an A+ student.
I was trying to graph x+1/x and x²+1/x² in my head while trying to get to sleep last night. I think I was thinking about particle forces and galaxies or something, but I think I fell asleep before I finished. I guess it's better than counting sheep anyway! But it was pretty weird seeing the same expression here on TH-cam when I woke up..
I just multiplied original equation by x taking note that x could not be zero and then factor the resulting quadratic to find that x has to be 1. Sub into second equation and boom, it equals 2.
This basic problem appears elsewhere properly stated it should be posed thus “ WITHOUT SOLVING for x find blah blah if blah” Erudite integral fans will remember that a variation of this problem appears in Nahin’s Inside Interesting Integrals
I think that maybe you should've told (edit: or tell, like in some other video?) more about substitution t=x+1/x. You used it with n instead of t in this video and did something with it, but you could tell people much more, especially together with formula for (a+b)^n, which greatly simplifies for (x+1/x)^n and is very useful for many problems! 🙂Or even for some integrals! ;-) Even here, we could say that t^3=(x^3+1/x^3)+3(x+1/x), thus x^3+1/x^3=t^3-3t, and then from formula for (a+b)^n t^5=(x^5+1/x^5)+5(x^3+1/x^3)+10(x+1/x), which gives us: x^5+1/x^5=t^5-5(t^3-3t)-10t = t^5-5t^3+5t So, we essentially get what you did, but x+1/x=t is useful in many occasions, like some integrals, some biquadratic equations and many other olympiad problems that are mixing powers of x and 1/x :-) So, having direct approach like this could be educational too.
what I did (although I prolly could have guess/checked and I'm dumb for not realizing that x was 1) was bring everything over to 1 side (x + 1/x - 2 = 0) "pull out" a 1/x -> 1/x(x^2 + 1 - 2x) = 0 Factored (1/x)(x-1)(x-1) = 0 and solved, x=1 The way you did it gave a general form, which was way way cooler
Well, for n=1 there is actually no real solution, because the dicriminant of the quadratic x^2-nx+1=0 is n^2-4 and it has to be non-negative for real solutions to exist. So n needs to be is in the interval (-inf, -2> sum
this is the chad way to solve it. None of that quadratic talk. And don’t you come talking about that 1+1=2 thing, no no that is worse than solving the quadratic
Applying Cauchy inequality, there is a quick answer: First prove x and 1/x >0: x and 1/x are the same sign. If x x + 1/x 0. Then apply Cauchy inequality for x and 1/x: x + 1/x >= 2 , equality happens when x=1/x=>x=1=> x5 + 1/x5 = 2.
I know it wasn't the point of the video but one can also solve this problem quite quickly by solving directly for x by either letting 2=2sqrt(x)/sqrt(x) which gives (sqrt(x))^2+(1/sqrt(x))^2-2sqrt(x)(1/sqrt(x))=0 (we can do this because it is clear x has to be positive in the first equation) =>(sqrt(x)-1/sqrt(x))^2=0 =>sqrt(x)=1/sqrt(x) =>x=1 or using the Arithmetic Mean-Geometric Mean Inequality which tells us x+1/x can only be equal to 2 if x=1/x (for positive x which, again, we have) Both not very good methods since they don't really generalise
Hi, interested in math competitions? If so, take a look at th-cam.com/video/rkzxdMFEEtw/w-d-xo.html and others in the Olympiad playlist. You will see a lot of tricky and somewhat complex problems to try. Hope to see you there.
When x is negative 1/x is also negative therefore x + 1/x = 2 implies x is positive. now AM>=GM therefore x+1/x >= 2. equality holds only if x = 1/x or x = 1. therefore x^5 + 1/x^5 = 2
There is one answer when n=2, but two answers for all n>2. For example, with n=3, both 2.61 and 0.38 satisfy the first equation x + 1/x = 3, so x^5 + 1/x^5 = 123 and ≈ 2.63. For n between 0 and 2 there are no rational roots to the first equation, so I'm not sure if 1 is an answer to the latter equation. Edit: Sorry, I was wrong about that. Somehow I performed 0.38^5 + 1/(0.38^5) incorrectly. Both = 123. Brain farts on a Saturday morning.
x + 1/x >= 2 (use Cauchy Schwarz inequality) => x will be minimum x^5 + 1/x^5 >= 2 ( still use Cauchy Schwarz inequality) because x is minimum so x^5 + 1/x^5 will be minimum too so the answer is 2
This is a nice video with an elegant solution. Yet, I don't want to spoil it, but I noticed an illegitimate operation. The squaring of x + 1/x = 2 was not done correctly. As a result, x = -1 is a valid solution to x^2 + 1/x^2 = 2, but is not a solution to the original x + 1/x = 2. Squares and sqare roots are tricky bastards.
For the general case *x + 1 / x = m in N \ {1}, s_n := x^n + x^{-n} = ?* the solution is: *s_n = y^n + y^{-n}, y = m / 2 + sqrt(m^2 / 4 - 1)* If you're comfortable with complex numbers, the same is true for any " *m in C* ". You can directly check it via quadratic formula, as many have mentioned, induction or (much more interesting and general) use "Newton's Identity". In the case of induction or "Newton's Identity", the only thing you need to notice is the recursion *s_{n+1} = (x + 1 / x) * s_n - s_{n-1}, s_0 = 2, s_1 = m*
What about x^500 + 1/x^500? No need for the hassle. Just multiply both sides with x and get a second degree polynomial. Conveniently a+b+c = 0 (where a b c are coefficients of the polynomial) which makes x = 1 or x = c/a = 1 😎
I havent watched the video, but you can take the derivative of x and 1/x, and you get two different derivatives, which means they change at separate rates, and in this case one always increases as the other decreases. That means for every increase or decrease in the value of x, there will be a nonlinear decrease or increase respectively in the value of 1/x, meaning x + 1/x will only ever equal a given value for one value of x, for positive real x values. Negative values are negative here, and im just ignoring complex numbers because ugh. Solving for the derivatives being the same value, and you get 1. So x=1 is the only real answer.
Hi, interested in math competitions? If so, take a look at th-cam.com/video/rkzxdMFEEtw/w-d-xo.html and others in the Olympiad playlist. You will see a lot of tricky and somewhat complex problems to try. Hope to see you there.
Hi, interested in math competitions? If so, take a look at th-cam.com/video/rkzxdMFEEtw/w-d-xo.html and others in Olympiad playlist. You will see a lot of tricky and somewhat complex problems to try. Hope to see you there.
*Obviously*, this is highly *trivial* The keen-eyed will see at first glance that x + (1/x) = 2 has a solution x=1, since plugging in 1 into x makes the statement true. Absolute *prodigies* (e.g. me) will recognize immediately that 1 is an idempotent in the real numbers, hence, plugging in 1 into the expression x^5 + (1/x^5) will trivially be the same as the prior expression, which was set equal to two. Therefore, father Flamme, I have earned myself that sweet red heart symbol at the bottom of my comment, especially as I was able to surpass all those simpletons rambling about multiplying by x and solving a quadratic, which, by the way, I am able to solve within seconds thanks to my baffling mathematical capabilities, but that is, as you, father Flamme, may expect, a highly nontrivial story for another day.
x+1/x = n let f(r) = x^r+1/x^r so f(1) = n (x^r+1/x^r)(x+1/x) = nf(r) = x^(r+1) + 1/x^r + x/x^r + x^r/x = f(r+1) + 1/x^(r-1) + x^(r-1) = f(r+1) + f(r-1) f(r+1) = n*f(r) - f(r-1) there's a generalized recursive formula if I haven't messed anything up, don't know how to get a non recursive formula, but I have a feeling there's some cancellation you can do by recursively plugging it into itself or something. Also might be a way to derive f(r+c) from f(r) and f(c) that could be more helpful
Why go through something so complicated when you could just solve the thing in the first place and get x=1 in about 3 steps, and then plug that solution in and note that since x=1, x^n + 1/x^n =2 for any n.
I am use a complex method (Multiply "square power of X + 1/X = 2" and "cube power of X +1/X =2"), and someone in the comment section solve it with quadratic formula (Of factorize? i don't know english well). emotional damage
Hi Jalma. If you want to see similar videos in math competitions, please consider th-cam.com/video/rkzxdMFEEtw/w-d-xo.html and other videos in the Olympiad playlist. Hope you enjoy 😊
@@animemoments7777 build everything from the foundations up. start from the number line if you have to, just make sure you have a solid foundation. The resources you should use depend on your level really.
*_Don't forget: 10% of my Ad and Merch Income this month will go to humanitarian help for Ukraine. So please make sure to watch the video in its entirety and to check out some of today's relevant links! :)_*
Train your Algebra Expertise by trying out Brilliant! =D brilliant.org/FlammableMaths
Support the channel by checking out Deez Nutz over on stemerch.eu/ ! :3
Engi Watch: stemerch.com/products/the-incredibly-unrigorous-engineering-watch?variant=40377075728562
Wuck. play.google.com/store/apps/details?id=org.flammablemaths.Wuck
Check out my newest video over on @Flammy's Wood ! =D th-cam.com/video/tTpjWePfK7o/w-d-xo.html
Shirt from the Video: stemerch.eu/collections/fugacity
You're creative keep going✨✨✨✨✨..
Sender: your brother (Baraa) from Palestine
🇵🇸🇺🇲
👍
is not x+1/x minimum at 2 with x = 1?
As a Ukrainian, I would ask to send money not for humanitarian aid, but for the army. The main donations go to humanitarian (several times more), although the only thing that can save cities with civilians is the army. For example, you can donate to the "turn back alive"(повернись живим) fund , or any other that sends money to the army or to hospitals.
I'm really impressed how you made 1 + 1 so interesting. XD
:DDD
I imagine that the intent was to demonstrate properties of exponentiation, because the initial equation is just the classic quadratic x^2-2x+1=0 in disguise, which only has solution = 1, so of course 1^m+1/(1^n)=2 for any powers m and n.
Can we just appreciate that jens makes great videos that only get a few thousand views and doesn’t give up?
*cries in 20 views*
who is jens
@@Tomaplen jens flau, papa flammy
@@Tomaplen Jens Fehlau is the host/presenter of the Flammable Maths youtube channel.
I love the method you use because you could easily solve for x and plug in the value. But in this way you get to the solution without needing to know what x is. It's mathemagical.
Yup! That was the point :) Also, this way you can easily generalize it!
@@PapaFlammy69 Hello , i like your videos check this out because there are two ways to disprove it, one when x€R and x€C. Let this equation be;
x²+x+1=0 (1) , x≠0
Then we divide (1) by x;
x+1+1/x=0
From the first (1) equation we solve for x+1;
x²+x+1=0=> x+1=-x² (2)
Now we have; x+1 +1/x=0 =>(2) -x²+1/x=0=> x²=1/x Now we multiply by x (x≠0) and he have;
x³=1=> x=1. But if x=1; (1)=> x²+x+1=0=> 1²+1+1=0=> 1+1+1=0=> 3=0.
That's the "problem" but there are two ways to disprove it. I Found out one but the other when x€C was a bit confusing. I would appreciate if you answered or make a small video. (I am fan of your work). Have a nice day!
Why not first multiply it with x, giving us x^2-2x+1=(x-1)^2=0, thus x=1, thus x^5+1/(x^5)=1+1=2
sure, you can do that too.
Yea this is exactly what I did
I think that is less interesting. Yeah, it solves the problem, but it is more beautiful to see the pattern that occurs and the mystery of solving the problem without the actual value of x.
Why go the easy route, when you enjoy doing much more algebruh instead
@@robinbfh5893 Because I enjoy simple and short solutions the most
I have managed to solve this for the general case. This will be a long one, so buckle up.
So, let's say we have a(1) = x + x^-1 = k (we'll have a detailed discussion about k later) and we want to find the value of a(n) = x^n + x^-n. We're well aware that if n = 0, then a(0) = x^0 + x^-0 = 2, and for n = 1, it's given.
Then, if we want to find a(n + 1), we first multiply (x + x^-1)(x^n + x^-n) = a(1)a(n) = x^(n + 1) + x^-(n+1) + x^(n - 1) + x^-(n - 1) = a(n + 1) + a(n - 1). We know a(1), it's k, so our recurrence relation becomes a(n + 1) = k*a(n) - a(n - 1).
This is really important, because we can now use familiar linear recurrence solving techniques, because this is a linear recurrence with constant coefficients (k is constant w.r.t. this equation), a second order one at that (because of a(n) and a(n-1)). We can use the characteristic polynomial, it will be t^2 - kt + 1 = 0 which gives us t = k/2 ± √((k/2)² - 1). If we want t to be real, then (k/2)² - 1 >= 0 or (k/2)² >= 1, so k >= 2 or k
Very educational, thank you for this comment!! ♥
Many thanks for this: I'd got to the same formula for a(n) in a more brute-force way. In answer to your conjecture, I'm fairly sure that if k is an integer, then a(n) is an integer too - because the odd powers of the surds in the two parts of the formula cancel, leaving only the even powers. I'm now wondering whether there's a fairly simple closed formula for a(n) in terms of powers of k with integer coefficients.
If you let x equal some complex number e^iy, then x + 1/x = 2cos(y) and x^n + 1/x^n = 2cos(ny). Then you’re rephrasing the question into trying to find 2cos(ny) in terms of 2cos(y), which can be found by doing the real part of the expansion (cosx + isinx)^n. You might then be able to generalise
That's what i want. I actually want a solution for x + 1/x = k, what would x^n + 1/x^n = In term of k and n??
@@itsphoenixingtime the general formula is as if x + 1/x = n, x^a + x^-a = n^a - 3n^(a-2) - n^(a - 3) - n^(a - 4) . . . . - n
@@rangaweerakkody1346 Looks pretty cool, I'll check this out.
I looked at the title and thought “well maybe if u give me my old algebra notes to review and a couple cups of coffee” lmao
xD
Me who observed the only solution is x=1 :
Math and Jens accent go brilliantly well
At first I use the same method as you did. But later on I found that using AM-GM inequality is much easier.
x+1/x≥2√x(1/x)
x+1/x≥2
From this,we know that x+1/x=2 when x=1.So x^5+1/x^5=2
The only problem is that, for AM-GM x>0 must be given. (you could consider a case where it is greater than 0 and less than 0 and then get a full condition as follows: 1. if x is greater than 0 then, x+1/x > 2 (or equal) and 2. if x is less than 0, x+1/x < -2 (or equal)
Took me way too long to realize this is just 1+1.
Plot twist: He is Preshtoad walker
Where did you get your blackboard? I’ve been looking for one for ages but it seems you can only buy them in bulk.
Papa Flammy I have an idea that might involve more than one person. I really love that incredibly unrigorious engineering watch, however I wouldnt buy it becuase I already have a smartwatch and I would switch that. Howwever, I think that you could make that(or even other maths watch faces) for smartwatches, people would probably pay for it. I know there are already some out there, but I think we could use more funny desings like seen on the engineering watch
Hmmmm, that is a great idea!
i saw that problem and immediately just thought to multiply the first equation by x. gives you a simple quadratic
It just struck me: you could use am-gm inequality to solve this. However you'll only get a solution in the set of positive real numbers......
x+1/x ≥ 2 that means that x= 1/x = 1
Then x^5 + 1/x^5 = 2
Edit: the answer cannot have a negative value because if x < 0 then 1/x < 0 and hence x + 1/x < 0
The way you solved this is really cool :D
Or you could square (x+1/x) two times getting x^4 power and multiplying it with (x+1/x)
When he says "final solution" with that accent. Passover observers shook.
I want the with LOSS of generality shirt so bad but it's not available
Are they not supposed to sleep? Or what i don't get it 0:04
I don't get it either
The picture depicts how they sleep, yes. That's the point and meme
Take the original equation, multiply all sides by x, switch it around, and we have x^2-2x+1=0, or (x-1)^2=0. So then x=1 and we can substitute this into the x^5 equation to get 2
1729, the Hardy-Ramanujan Number, is the smallest number which can be expressed as the sum of two different cubes in two different ways.
Blessed is the land where Shrinivas Ramanujan were born in India.
Actually only if both the numbers are natural numbers if integers are consider 91 is the smallest number which can be expressed as sum of two cubes in two ways that is : 3³+4³=6³+(-5)³ .
Love the problems you post.
The first problem on this channel that I was able to solve the answer is 2 .Thanks flammable maths for this nice problem
This is boundary value problem ,
minimum positive value of x + 1/x is 2 which can be proved by am gm inequality so the value of x is 1 then put in x^5 + 1/x^5 which will give 1+ 1 = 2
With you, Of course I can solve it Dear Papa ...
Let's solve it
I received a very similar question to this from SMO senior a little while before I saw this vid. I was able to solve it using your technique. thanks papa!
Hi, if interested in math competitions (including SMO)? If so, take a look at th-cam.com/video/rkzxdMFEEtw/w-d-xo.html and others in the Olympiad playlist. You will see a lot of tricky and somewhat complex problems to try. Hope to see you there.
I am not well versed in the formalism of mathematics and idk if the following is valid.
Generalize this problem to the complex world, hence
z+1/z=2
if we limit ourselves to the unit circle in the complex plane, we get r = 1. [idk if we can do this or not without losing the fact that x is not bounded]
2cos(a)=2 -> a=2*pi*k where k∈Z
for the 5th case
z^5 + z^(-5)
2cos(5a) = 2
hence x^5 + 1/x^5 = 2 ?
again, I have no idea about formalisms and errors of assuming it to be in a restricted part of the complex plane. But it works?
You can direct take power 5 on both sides the expand binomially
we can just use cauchy inequality for this also! x + 1/x >= 2√x.1/x = 2, and we also hav x + 1/x = 2 => x = 1/x => x = 1 so we can lead that to the final answer !
The graph for x+(1/x) is very cool check it out
Here in india we are taught how to find the range of several functions
And this is one of the infamous ones
It's range goes as
(-infinity,-2] U [2,Inf)
Where -2 and 2 occur and
-1 and 1 respectively
I personally just solved for x and subbed in, but your way of solution really got me thinking. Guess it is the difference between an A student and an A+ student.
I was trying to graph x+1/x and x²+1/x² in my head while trying to get to sleep last night. I think I was thinking about particle forces and galaxies or something, but I think I fell asleep before I finished.
I guess it's better than counting sheep anyway!
But it was pretty weird seeing the same expression here on TH-cam when I woke up..
good job making a no brainer so fascinating xD
I didn't get the meme at the start, what was that
I just multiplied original equation by x taking note that x could not be zero and then factor the resulting quadratic to find that x has to be 1. Sub into second equation and boom, it equals 2.
I found out from x+1/x=2 that x2+1/x2 is also =2 so i thought it could probably only be possible when x=1
0:09 is >mfw realize the type of problem we're about to look at in this video
This basic problem appears elsewhere properly stated it should be posed thus
“ WITHOUT SOLVING for x find blah blah if blah”
Erudite integral fans will remember that a variation of this problem appears in Nahin’s Inside Interesting Integrals
the general formula is as if x + 1/x = n, x^a + x^-a = n^a - 3n^(a-2) - n^(a - 3) - n^(a - 4) . . . . - n
Is there a general solution (e.g. proof by induction) for x^n add x^(-n). Intuitively it could be 2 as well.
I think that maybe you should've told (edit: or tell, like in some other video?) more about substitution t=x+1/x. You used it with n instead of t in this video and did something with it, but you could tell people much more, especially together with formula for (a+b)^n, which greatly simplifies for (x+1/x)^n and is very useful for many problems! 🙂Or even for some integrals! ;-)
Even here, we could say that
t^3=(x^3+1/x^3)+3(x+1/x), thus x^3+1/x^3=t^3-3t,
and then from formula for (a+b)^n
t^5=(x^5+1/x^5)+5(x^3+1/x^3)+10(x+1/x),
which gives us:
x^5+1/x^5=t^5-5(t^3-3t)-10t = t^5-5t^3+5t
So, we essentially get what you did, but x+1/x=t is useful in many occasions, like some integrals, some biquadratic equations and many other olympiad problems that are mixing powers of x and 1/x :-)
So, having direct approach like this could be educational too.
Another commenter mentioning the cosine function makes me think there could be some Chebyshev polynomial action going on.
what I did (although I prolly could have guess/checked and I'm dumb for not realizing that x was 1) was
bring everything over to 1 side (x + 1/x - 2 = 0)
"pull out" a 1/x -> 1/x(x^2 + 1 - 2x) = 0
Factored (1/x)(x-1)(x-1) = 0
and solved, x=1
The way you did it gave a general form, which was way way cooler
cant you also use some forms of inequalities to solve for x which is equal to 1 then put it back to the og equation?
Omg I finally got the answer on the problem for once!
Geometry Euclides please 🥺🥺
Or you could solve for x: since X is in the denominator x != 0, and then x²-2x+1=(x-1)²=0=x-1 and X=1. Then x^5 + x^-5 = 2
This method does not generalize, though.
@@angelmendez-rivera351 It actually does generalize. You solve a quadratic and whatever solutions you choose give the same answer.
@@muriloporfirio7853 Yes, but that has to be proven, and does not follow non-trivially from what you did. Hence, my point stands.
Well, for n=1 there is actually no real solution, because the dicriminant of the quadratic x^2-nx+1=0 is n^2-4 and it has to be non-negative for real solutions to exist. So n needs to be is in the interval (-inf, -2> sum
Why should it be real though?
i used newton biname
If I got x=1 so ? = 2 from the thumbnail am I smart or dumb?
this is the chad way to solve it.
None of that quadratic talk.
And don’t you come talking about that 1+1=2 thing, no no that is worse than solving the quadratic
Applying Cauchy inequality, there is a quick answer: First prove x and 1/x >0: x and 1/x are the same sign. If x x + 1/x 0. Then apply Cauchy inequality for x and 1/x: x + 1/x >= 2 , equality happens when x=1/x=>x=1=> x5 + 1/x5 = 2.
I know it wasn't the point of the video but one can also solve this problem quite quickly by solving directly for x by either letting 2=2sqrt(x)/sqrt(x) which gives (sqrt(x))^2+(1/sqrt(x))^2-2sqrt(x)(1/sqrt(x))=0 (we can do this because it is clear x has to be positive in the first equation)
=>(sqrt(x)-1/sqrt(x))^2=0
=>sqrt(x)=1/sqrt(x)
=>x=1
or using the Arithmetic Mean-Geometric Mean Inequality which tells us x+1/x can only be equal to 2 if x=1/x (for positive x which, again, we have)
Both not very good methods since they don't really generalise
It's amazing 🤯🤯 It is possible to evaluate the expression without looking for the specific value of x 🤯🤯🤯🤯🤯
You're creative keep going✨✨✨✨✨..
Sender: your brother (Baraa) from Palestine
🇵🇸🇺🇲
I don’t get the meme in the beginning
Thank u Sir.
i dont get the italian bit
Flammerble Meffs posted! 🐐🥛
I just solved for x at x+1/x=2 and got 1 then substituted 1 at the x⁵+1/x⁵=2
Right or wrong lol
Will you make some more videos concerning more advanced maths, like complex analysis or some kind of integral?
I just x+1/x=2 -> x^2-2x+1=0 the quadratic has 1 root and its 1. Why do it the hard way?
Hi, interested in math competitions? If so, take a look at th-cam.com/video/rkzxdMFEEtw/w-d-xo.html and others in the Olympiad playlist. You will see a lot of tricky and somewhat complex problems to try. Hope to see you there.
When x is negative 1/x is also negative therefore x + 1/x = 2 implies x is positive. now AM>=GM therefore x+1/x >= 2. equality holds only if x = 1/x or x = 1. therefore x^5 + 1/x^5 = 2
This has some Wau number vid vibes...
I assumed the equation as K and squared both sides, then applied A.M. G.M. inequality to get K² = 4 answer found
There is one answer when n=2, but two answers for all n>2. For example, with n=3, both 2.61 and 0.38 satisfy the first equation x + 1/x = 3, so x^5 + 1/x^5 = 123 and ≈ 2.63. For n between 0 and 2 there are no rational roots to the first equation, so I'm not sure if 1 is an answer to the latter equation.
Edit: Sorry, I was wrong about that. Somehow I performed 0.38^5 + 1/(0.38^5) incorrectly. Both = 123. Brain farts on a Saturday morning.
41* 3 is 123
Hence 123 is not a prime no. Bcoz it is divisible by 3
x + 1/x >= 2 (use Cauchy Schwarz inequality) => x will be minimum
x^5 + 1/x^5 >= 2 ( still use Cauchy Schwarz inequality)
because x is minimum so x^5 + 1/x^5 will be minimum too so the answer is 2
Hypothesis means x=1so
This is a nice video with an elegant solution. Yet, I don't want to spoil it, but I noticed an illegitimate operation.
The squaring of x + 1/x = 2 was not done correctly.
As a result, x = -1 is a valid solution to x^2 + 1/x^2 = 2, but is not a solution to the original x + 1/x = 2.
Squares and sqare roots are tricky bastards.
The squaring was done correctly. x=-1 is simply an extraneous solution.
x+1/x≥2, equal 2 only then x=1
If you are restricting to only reals, then x+1/x = n := 1 is unsolvable, since x must be complex.
For the general case
*x + 1 / x = m in N \ {1}, s_n := x^n + x^{-n} = ?*
the solution is:
*s_n = y^n + y^{-n}, y = m / 2 + sqrt(m^2 / 4 - 1)*
If you're comfortable with complex numbers, the same is true for any " *m in C* ". You can directly check it via quadratic formula, as many have mentioned, induction or (much more interesting and general) use "Newton's Identity". In the case of induction or "Newton's Identity", the only thing you need to notice is the recursion
*s_{n+1} = (x + 1 / x) * s_n - s_{n-1}, s_0 = 2, s_1 = m*
What is wrong with this method?
1) X + 1/X = 2
2) (X/X)*X + 1/X = 2
3) (X^2 + 1)/X = 2
4) X*(X^2 + 1)/X = 2*X
5) X^2 + 1 = 2X
6) X^2 -2x + 1 = 0
7) (X - 1) * (X - 1) = 0 Therefore, X = 1
So, 1^5 + 1/1^5 = 2
Fun fact: X + 1/X >= 2 for all positive real numbers. It is equal to 2 in only one case ( x = 1 ).
i saw the thumbnail and wondered how it was even a question lol
What about x^500 + 1/x^500? No need for the hassle. Just multiply both sides with x and get a second degree polynomial. Conveniently a+b+c = 0 (where a b c are coefficients of the polynomial) which makes x = 1 or x = c/a = 1 😎
a method only for concluding a correct answer is special results usually occur in special cases.
here is x＝1
Lol. Take x+l/x=2 time it with x and we got quadratic, x^2+1=2x. So x=1. So x^5+1/x^5=2
I havent watched the video, but you can take the derivative of x and 1/x, and you get two different derivatives, which means they change at separate rates, and in this case one always increases as the other decreases. That means for every increase or decrease in the value of x, there will be a nonlinear decrease or increase respectively in the value of 1/x, meaning x + 1/x will only ever equal a given value for one value of x, for positive real x values. Negative values are negative here, and im just ignoring complex numbers because ugh. Solving for the derivatives being the same value, and you get 1. So x=1 is the only real answer.
Hi, interested in math competitions? If so, take a look at th-cam.com/video/rkzxdMFEEtw/w-d-xo.html and others in the Olympiad playlist. You will see a lot of tricky and somewhat complex problems to try. Hope to see you there.
I mean, you could've solved the quadratic equation first and plug in into the second relation...
Hi, interested in math competitions? If so, take a look at th-cam.com/video/rkzxdMFEEtw/w-d-xo.html and others in Olympiad playlist. You will see a lot of tricky and somewhat complex problems to try. Hope to see you there.
You said "induction"
I know what I must do o.o
Unless you do it xD
*Obviously*, this is highly *trivial*
The keen-eyed will see at first glance that x + (1/x) = 2 has a solution x=1, since plugging in 1 into x makes the statement true.
Absolute *prodigies* (e.g. me) will recognize immediately that 1 is an idempotent in the real numbers, hence, plugging in 1 into the expression x^5 + (1/x^5) will trivially be the same as the prior expression, which was set equal to two.
Therefore, father Flamme, I have earned myself that sweet red heart symbol at the bottom of my comment, especially as I was able to surpass all those simpletons rambling about multiplying by x and solving a quadratic, which, by the way, I am able to solve within seconds thanks to my baffling mathematical capabilities, but that is, as you, father Flamme, may expect, a highly nontrivial story for another day.
great!!
x+1/x = n
let f(r) = x^r+1/x^r
so f(1) = n
(x^r+1/x^r)(x+1/x) = nf(r)
= x^(r+1) + 1/x^r + x/x^r + x^r/x
= f(r+1) + 1/x^(r-1) + x^(r-1)
= f(r+1) + f(r-1)
f(r+1) = n*f(r) - f(r-1)
there's a generalized recursive formula if I haven't messed anything up, don't know how to get a non recursive formula, but I have a feeling there's some cancellation you can do by recursively plugging it into itself or something.
Also might be a way to derive f(r+c) from f(r) and f(c) that could be more helpful
Holy i learned this in school
Why go through something so complicated when you could just solve the thing in the first place and get x=1 in about 3 steps, and then plug that solution in and note that since x=1, x^n + 1/x^n =2 for any n.
I am use a complex method (Multiply "square power of X + 1/X = 2" and "cube power of X +1/X =2"), and someone in the comment section solve it with quadratic formula (Of factorize? i don't know english well). emotional damage
Hi Jalma. If you want to see similar videos in math competitions, please consider
th-cam.com/video/rkzxdMFEEtw/w-d-xo.html and other videos in the Olympiad playlist. Hope you enjoy 😊
You can simply solve for the value of x and proceed from there, but that's not how real men do math.
Indeed.
You can see with the binom formula that x has to be 1 to fulfill the first equation.
So its pretty obvious.😉
I can't
nice.
@@PapaFlammy69 Bro how can I learn maths on my own? My teacher don't give a sht about my class and I don't have anyone to guide me
@@animemoments7777 khanacademy, math textbooks, math videos etc.
@@animemoments7777 build everything from the foundations up. start from the number line if you have to, just make sure you have a solid foundation. The resources you should use depend on your level really.
to answer your question, no.
@11:34 the meaning of life. why is this joke so famous.
Because it's the one and only answer to life, the universe and everything
Why can't you just solve the first one as a quadratic equation?
Can You Figure Out this Algebra Problem?
Yes.
Wow! 1+1=2 ! Who could’ve known!
I figured it out through binomial formula.
(Long silence)
I'm ashamed of myself.
Лол это же неравенство коши. x+1/x>=2 ну и рассмотреть на положительных и отрицательных. Ответ ±2
А пардон минус 1 не то