x^65 + 1/x^65 can be easily expressed in terms of x + 1/x. Notice that x^0 + 1/x^0 = 2, so if we consider g(n) = x^n + 1/x^n, then g(0) = 2, and given g(1), we want to find g(65) as a function of g(1). We know that (x + 1/x)·[x^(n + 1) + 1/x^(n + 1)] = x^(n + 2) + 1/x^(n + 2) + x^n + 1/x^n, which is equivalent to g(1)·g(n + 1) = g(n + 2) + g(n), which has characteristic polynomial t^2 - g(1)·t + 1 = 0, which is equivalent to [t - g(1)/2]^2 + 1 - [g(1)/2]^2, which is equivalent to [t - g(1)/2]^2 = [g(1)/2]^2 - 1], which is equivalent to t = g(1)/2 - sqrt([g(1)/2]^2 - 1) or t = g(1)/2 + sqrt([g(1)/2]^2 - 1). Therefore, g(n) = A·{g(1)/2 - sqrt([g(1)/2]^2 - 1)}^n + B·{g(1)/2 + sqrt([g(1)/2]^2 - 1)}^n, and using the initial conditions, it can be proven that A = B = 1, hence g(n) = {g(1)/2 - sqrt([g(1)/2]^2 - 1)}^n + {g(1)/2 + sqrt([g(1)/2]^2 - 1)}^n. Let g(1) = 1, hence g(1)/2 - sqrt([g(1)/2]^2 - 1) = 1/2 - sqrt(1/4 - 1) = 1/2 - sqrt(3)/2·i = cos(π/3) - sin(π/3)·i = exp(-π/3·i), while g(1)/2 + sqrt([g(1)/2]^2 - 1) = exp(π/3·i), so g(n) = exp(n·π·i/3) + exp(-n·π·i/3) = cos(n·π/3) + sin(n·π/3)·i + cos(n·π/3) - sin(n·π/3)·i = 2·cos(n·π/3). If T[n, u] is the nth Chebyshev polynomial of order n, then T[n, cos(t)] = cos(n·t), hence g(n) = 2·cos(n·π/3) = 2·T[n, cos(π/3)] = 2·T[n, 1/2], so f(1) = 2·T[65, 1/2] = 2·cos(65·π/3). 65·π/3 = (66 - 1)·π/3 = 22·π - π/3 = 11·(2·π) - π/3, so 2·cos(65·π/3) = 2·cos(-π/3) = 2·cos(π/3) = 2·1/2 = 1. Therefore, f(1) = 1, and in fact, more generally, f(x) = (x/2 + sqrt[(x/2) - 1])^65 + (x/2 - sqrt[(x/2) - 1])^65, thus solving the functional equation over C.
@@anshumanagrawal346 For the purposes of this exercise, it actually does not matter, since all choices of branch give the same uniquely well-defined f. This is because if you replace all radicals with their negative, you get the same expression.
In the first one, you could skip some steps. You had x^4 = -x. Raising each side to the fourth power would give you x^16 = x^4, so x^16 = -x. Raising each side to the 4th power again, you get x^64 = x^4, giving you x^64 = -x again.
There is quite a handy formula which is related to the 2nd method. If z = cosθ + i sinθ, then for all integers n: z^n + 1/z^n = 2cos(nθ) You can use this to find the argument of a number x that satisfies x + 1/x = 1, and from there to find the value of x^65 + 1/x^65.
What an interesting function. Just doing it in my head, but I think the answer is... 1? I believe that the function is real-valued everywhere (a surprise - I didn’t think it would be at first.)
I'm not sure what function you are referring to when you say "the function is real-valued everywhere". If you mean the function x+1/x, this is real-valued for complex x precisely when x is real and non-zero or has modulus 1. Here are three proofs. 1st proof Suppose x≠0, and let x=a+bi, with a, b real. Then x+1/x=a+bi + 1/(a+bi) = a+bi + (a-bi)/[(a+bi)(a-bi)] (as a-bi≠0) = a+bi + (a-bi)/(a²+b²) = [(a+bi)(a²+b²) + (a-bi)]/(a²+b²) = [a(a²+b²+1) + (a²+b²-1)bi]/(a²+b²) The imaginary part (a²+b²-1)b/(a²+b²) is zero iff (a²+b²-1)b=0 a²+b²-1=0 or b=0 a²+b²=1 or b=0 |x|²=1 or b=0 |x|=1 (as |x|>0) or b=0 i.e. x has modulus 1 or x is real and non-zero. 2nd proof Suppose x≠0, and let x=re^(iθ), where r>0 and θ real, so 1/x=r⁻¹e^(-iθ), and x+1/x=re^(iθ)+r⁻¹e^(-iθ) =r(cosθ+isinθ)+r⁻¹(cosθ-isinθ) =(r+r⁻¹)cosθ+i(r-r⁻¹)sinθ. The imaginary part (r-r⁻¹)sinθ=0 iff r-r⁻¹=0 or sinθ=0. r-r⁻¹=0 r²-1=0 r=1 (as r>0), i.e. x has modulus 1 and sinθ=0 x is real and non-zero.. 3rd proof A complex number is real iff it equals its complex conjugate. So, for complex non-zero x, x+1/x is real iff x+1/x=(x+1/x)*=x*+(1/x)*=x*+1/x* (x-x*)+(1/x-1/x*)=0 (x-x*)+(x*-x)/(xx*)=0 (x-x*)+(x*-x)/|x|²=0 |x|²(x-x*)+(x*-x)=0 (as |x|²≠0) (|x|²-1)(x-x*)=0 |x|²-1=0 or x-x*=0 |x|=1 (as |x|>0) or x=x* x has modulus 1 or x is real and non-zero.
@@MichaelRothwell1 also got this conclusion using conjugates. Actually, it also applies to the function x^n + 1/x^n. It works because (z^n)*=(z*)^n, and in the end you'll arrive at z*^n • z^n = (z•z*)^n = 1. Difference is only that you'll have also additional real outputs for complex non-unit-module input x with 2n different arguments theta = (π/n)*k, k - arbitrary integer. That's because [r•e^(iπk/n)]^n leads to real values too even with r≠1. But I guess what might be meant is this. The thing is if you have an equation f(x+ 1/x) = x^n + 1/x^n, then resultant f(x) turns out to be real for any real n and x (not only "1" as in this example). Check it out. Substitute x + 1/x = t, then you get that x = t/2 ± √(t^2 - 4)/2. If t≥2, x is real, and RHS expression is obviously real - trivial. But if t
Nice problem with nice multiple solutions as usual. Here is a variant on Method 3: Show x³=-1 as before. Then x⁶=1, x⁶⁶=(x⁶)¹¹=1¹¹=1, x⁶⁵=1/x, x⁶⁵+1/x⁶⁵=1/x+x=1.
abbreviation: for real x with |x|>=2 let a := (x+sqrt(x²-4))/2 and b := (x-sqrt(x²-4))/2; then obviously a*b = 1 or a = 1/b, and a + 1/a = a + b = x. Now let the function f_n: R -> R be defined as follows (n is any natural number) : f_n(x) := 0 for |x| < 2 f_n(x) := a^n + b^n for |x| >=2 then for all real y > 0 we define x := y + 1/y and a,b as above; note |x| >=2. by construction we know for y>1 or -1
You can also write the solution of x^2 - x + 1 = 0 as -ω and -ω^2 from complex number analysis. Now, using the fact that ω = 1/ω^2 and 1 + ω + ω^2 = 0. The answer is simply solved to one
Actually in method 1 one of the steps was x^4=-x Since x is non zero divide both sides by x so you get x^3=-1 and the rest would be the same as method 3. Also x^65+1/x^65 = -(x^2+1/x^2) = -((x+1/x)^2-2) =1
Thanks for this interesting function. f(x+1/x)=x^65+1/x^65 I went further in the study of it. I have been looking for an explicit expression of f. Explicit expression ? Suppose y=x+1/x (looking f(y) expression) then x^2-yx+1=0 and x solves in 2 cases: -Case 1: -22 then x=(y+/-sqrt(y^2-4))/2 So case 1 gives an x of module as |x|^2=1/4(y^2+(4-y^2))=4/4=1 then module of x=1 Let a=arg(x), then cos(a)=y/2/1=y/2 And a=arg(x)=arccos(y/2) x can then be written as e^+/-ia so f(y) is as follows: f(y)=e^+i65a+e^-i65a with a=arg(x)=arccos(y/2) Finally, for -2
In this problem, the absolute value of x is 1, and the order of the expression to be obtained is as high as 65, so the second method-- x is expressed in the polar form of trigonometric functions, and De Moivre's theorem is used on the unit circle of the complex plane-- is the easiest and easiest to understand. I think you should get used to this method as it can be applied to similar problems. x+1/x=1 ⇒ x^2-x+1=0 ⇒ x=(1±i√3)/2, and expressed in polar form of trigonometric function, It becomes x=cos(±π/3)+i*sin(±π/3). Using De Moivre's theorem based on this, They will be x^6={cos(±π/3)+i*sin(± π/3)}^6=cos(±2π)+i*sin(±2π)=1, x^3={cos(±π/3)+i * sin(±π/3)}^3=cos(± π)+i*sin(±π)=-1, x^2= cos(±π/3)+i*sin(±π/3)}^2=cos(±2π/3)+i*sin(±2π/3) =-1/2±i√3/2. In other words, from the initial line, 6th power is 1 rotation, cubed is half rotation, and squared is 2π/3(120 °) rotation. Therefore, It can be solved as x^65={(x^6)^10}*(x^3)*(x^2)={(1)^10}*(-1)*(x^2)=(1)*(-1)* (x^2)=-(x^2), therefore, x^65+1/(x^65)=-(x^2)-1/(x^2) =-{( x ^2)+1/(x^2)}=-{(-1/2+i√3/2)+(-1/2-i√3/2)} or -{(-1/2)-i√3/2)+(-1/2+i√3/2)}=-(-1/2-1/2) =-(-1) = 1. No matter how much the multiplier of x increases, if you can understand that the value "turns on the unit circle", then all you have to do is know which position [1+i*0, 1/2+i*(√3/2),-1/2+i*(√3/2), -1+i*0, -1/2-i *(√3/2), 1/2-i*(√3/2)] to go by the multiplier.
x+1/x is continuous on the set of real numbers excluding 0..consider the sequence Xn(say) (1+1/n) converging to 1 so f(Xn) converges to f(1).. So limit(n tends to ♾️) f(Xn)=lim(n tends to ♾️)(1^65+(1/n)^65)=1=f(1).. So f(1)=1..
Hi. Marvelous video) I want to ask you about problems, where did you find such interesting themes and examples? You explored it in the internet or there are cool books? Pls recommend😊
Thank you! Books, internet, friends and myself... This one I made up (something that I enjoy doing) but that's easy to do if you've dealt with these kinds of equations before.
x^65 + 1/x^65 can be easily expressed in terms of x + 1/x. Notice that x^0 + 1/x^0 = 2, so if we consider g(n) = x^n + 1/x^n, then g(0) = 2, and given g(1), we want to find g(65) as a function of g(1). We know that (x + 1/x)·[x^(n + 1) + 1/x^(n + 1)] = x^(n + 2) + 1/x^(n + 2) + x^n + 1/x^n, which is equivalent to g(1)·g(n + 1) = g(n + 2) + g(n), which has characteristic polynomial t^2 - g(1)·t + 1 = 0, which is equivalent to [t - g(1)/2]^2 + 1 - [g(1)/2]^2, which is equivalent to [t - g(1)/2]^2 = [g(1)/2]^2 - 1], which is equivalent to t = g(1)/2 - sqrt([g(1)/2]^2 - 1) or t = g(1)/2 + sqrt([g(1)/2]^2 - 1). Therefore, g(n) = A·{g(1)/2 - sqrt([g(1)/2]^2 - 1)}^n + B·{g(1)/2 + sqrt([g(1)/2]^2 - 1)}^n, and using the initial conditions, it can be proven that A = B = 1, hence g(n) = {g(1)/2 - sqrt([g(1)/2]^2 - 1)}^n + {g(1)/2 + sqrt([g(1)/2]^2 - 1)}^n. Let g(1) = 1, hence g(1)/2 - sqrt([g(1)/2]^2 - 1) = 1/2 - sqrt(1/4 - 1) = 1/2 - sqrt(3)/2·i = cos(π/3) - sin(π/3)·i = exp(-π/3·i), while g(1)/2 + sqrt([g(1)/2]^2 - 1) = exp(π/3·i), so g(n) = exp(n·π·i/3) + exp(-n·π·i/3) = cos(n·π/3) + sin(n·π/3)·i + cos(n·π/3) - sin(n·π/3)·i = 2·cos(n·π/3). If T[n, u] is the nth Chebyshev polynomial of order n, then T[n, cos(t)] = cos(n·t), hence g(n) = 2·cos(n·π/3) = 2·T[n, cos(π/3)] = 2·T[n, 1/2], so f(1) = 2·T[65, 1/2] = 2·cos(65·π/3). 65·π/3 = (66 - 1)·π/3 = 22·π - π/3 = 11·(2·π) - π/3, so 2·cos(65·π/3) = 2·cos(-π/3) = 2·cos(π/3) = 2·1/2 = 1. Therefore, f(1) = 1, and in fact, more generally, f(x) = (x/2 + sqrt[(x/2) - 1])^65 + (x/2 - sqrt[(x/2) - 1])^65, thus solving the functional equation over C.
Very nice, as always!!! 🧡
How is sqrt defined here?
@@anshumanagrawal346 For the purposes of this exercise, it actually does not matter, since all choices of branch give the same uniquely well-defined f. This is because if you replace all radicals with their negative, you get the same expression.
@@angelmendez-rivera351 Interesting
@@angelmendez-rivera351 But still, in general how is the square root of a complex number defined? In my class we just learned it to have both valued
Nice!! I know how to solve it by complex numbers method x=e^ipi/3,but your methods are great!😀💯
That's right. Your way is more elegant.
Thank you for your videos - while RUclips takes 30% as fees, patreon will only take 8%. You might want to do that
Good idea!
In the first one, you could skip some steps. You had x^4 = -x. Raising each side to the fourth power would give you x^16 = x^4, so x^16 = -x. Raising each side to the 4th power again, you get x^64 = x^4, giving you x^64 = -x again.
Good thinking!
I'm very grateful you for this video and for your playlist about functional equations.
Np. Glad to hear that! 🥰
There is quite a handy formula which is related to the 2nd method.
If z = cosθ + i sinθ, then for all integers n:
z^n + 1/z^n = 2cos(nθ)
You can use this to find the argument of a number x that satisfies x + 1/x = 1, and from there to find the value of x^65 + 1/x^65.
Nice!
The last method was AWESOME!
Thank you! 🤩🤩
What an interesting function.
Just doing it in my head, but I think the answer is...
1?
I believe that the function is real-valued everywhere (a surprise - I didn’t think it would be at first.)
I'm not sure what function you are referring to when you say "the function is real-valued everywhere".
If you mean the function x+1/x, this is real-valued for complex x precisely when x is real and non-zero or has modulus 1.
Here are three proofs.
1st proof
Suppose x≠0, and let x=a+bi, with a, b real.
Then x+1/x=a+bi + 1/(a+bi)
= a+bi + (a-bi)/[(a+bi)(a-bi)] (as a-bi≠0)
= a+bi + (a-bi)/(a²+b²)
= [(a+bi)(a²+b²) + (a-bi)]/(a²+b²)
= [a(a²+b²+1) + (a²+b²-1)bi]/(a²+b²)
The imaginary part (a²+b²-1)b/(a²+b²) is zero iff (a²+b²-1)b=0
a²+b²-1=0 or b=0
a²+b²=1 or b=0
|x|²=1 or b=0
|x|=1 (as |x|>0) or b=0
i.e. x has modulus 1 or x is real and non-zero.
2nd proof
Suppose x≠0, and let x=re^(iθ), where r>0 and θ real,
so 1/x=r⁻¹e^(-iθ),
and x+1/x=re^(iθ)+r⁻¹e^(-iθ)
=r(cosθ+isinθ)+r⁻¹(cosθ-isinθ)
=(r+r⁻¹)cosθ+i(r-r⁻¹)sinθ.
The imaginary part (r-r⁻¹)sinθ=0 iff r-r⁻¹=0 or sinθ=0.
r-r⁻¹=0 r²-1=0 r=1 (as r>0), i.e. x has modulus 1
and sinθ=0 x is real and non-zero..
3rd proof
A complex number is real iff it equals its complex conjugate.
So, for complex non-zero x,
x+1/x is real iff
x+1/x=(x+1/x)*=x*+(1/x)*=x*+1/x*
(x-x*)+(1/x-1/x*)=0
(x-x*)+(x*-x)/(xx*)=0
(x-x*)+(x*-x)/|x|²=0
|x|²(x-x*)+(x*-x)=0 (as |x|²≠0)
(|x|²-1)(x-x*)=0
|x|²-1=0 or x-x*=0
|x|=1 (as |x|>0) or x=x*
x has modulus 1 or x is real and non-zero.
@@MichaelRothwell1 Wow!
How could it be? That would contradict the fact that Polynomial Equations are always solvable in C
@@MichaelRothwell1 also got this conclusion using conjugates. Actually, it also applies to the function x^n + 1/x^n. It works because (z^n)*=(z*)^n, and in the end you'll arrive at z*^n • z^n = (z•z*)^n = 1. Difference is only that you'll have also additional real outputs for complex non-unit-module input x with 2n different arguments theta = (π/n)*k, k - arbitrary integer. That's because [r•e^(iπk/n)]^n leads to real values too even with r≠1.
But I guess what might be meant is this. The thing is if you have an equation f(x+ 1/x) = x^n + 1/x^n, then resultant f(x) turns out to be real for any real n and x (not only "1" as in this example). Check it out. Substitute x + 1/x = t, then you get that x = t/2 ± √(t^2 - 4)/2. If t≥2, x is real, and RHS expression is obviously real - trivial. But if t
Did you want to say "real-valued everywhere in R"? Since f(i) already returns two different complex values
Nice problem with nice multiple solutions as usual.
Here is a variant on Method 3:
Show x³=-1 as before.
Then x⁶=1, x⁶⁶=(x⁶)¹¹=1¹¹=1,
x⁶⁵=1/x, x⁶⁵+1/x⁶⁵=1/x+x=1.
Very good!
x^3+1=(x+1)*(x^2-x+1). Q.e.d
SyberMaths love for functional equation will never get old
That's true! I'm in love with functions!!!
abbreviation: for real x with |x|>=2 let a := (x+sqrt(x²-4))/2 and b := (x-sqrt(x²-4))/2; then obviously a*b = 1 or a = 1/b, and a + 1/a = a + b = x.
Now let the function f_n: R -> R be defined as follows (n is any natural number) :
f_n(x) := 0 for |x| < 2
f_n(x) := a^n + b^n for |x| >=2
then for all real y > 0 we define x := y + 1/y and a,b as above; note |x| >=2.
by construction we know
for y>1 or -1
You can also write the solution of x^2 - x + 1 = 0 as -ω and -ω^2 from complex number analysis. Now, using the fact that ω = 1/ω^2 and 1 + ω + ω^2 = 0. The answer is simply solved to one
Actually in method 1 one of the steps was x^4=-x
Since x is non zero divide both sides by x so you get x^3=-1 and the rest would be the same as method 3.
Also x^65+1/x^65 = -(x^2+1/x^2) = -((x+1/x)^2-2) =1
Thanks for this interesting function.
f(x+1/x)=x^65+1/x^65
I went further in the study of it.
I have been looking for an explicit expression of f.
Explicit expression ?
Suppose y=x+1/x (looking f(y) expression)
then x^2-yx+1=0 and x solves in 2 cases:
-Case 1: -22 then x=(y+/-sqrt(y^2-4))/2
So case 1 gives an x of module as
|x|^2=1/4(y^2+(4-y^2))=4/4=1 then module of x=1
Let a=arg(x), then cos(a)=y/2/1=y/2
And a=arg(x)=arccos(y/2)
x can then be written as e^+/-ia so f(y) is as follows:
f(y)=e^+i65a+e^-i65a with a=arg(x)=arccos(y/2)
Finally, for -2
Wow!!!
Please throw some questions on recursive functional equations and its solutions thru characteristics equations
In this problem, the absolute value of x is 1, and the order of the expression to be obtained is as high as 65, so the second method-- x is expressed in the polar form of trigonometric functions, and De Moivre's theorem is used on the unit circle of the complex plane-- is the easiest and easiest to understand. I think you should get used to this method as it can be applied to similar problems. x+1/x=1 ⇒ x^2-x+1=0 ⇒ x=(1±i√3)/2, and expressed in polar form of trigonometric function, It becomes x=cos(±π/3)+i*sin(±π/3). Using De Moivre's theorem based on this, They will be x^6={cos(±π/3)+i*sin(± π/3)}^6=cos(±2π)+i*sin(±2π)=1, x^3={cos(±π/3)+i * sin(±π/3)}^3=cos(± π)+i*sin(±π)=-1, x^2= cos(±π/3)+i*sin(±π/3)}^2=cos(±2π/3)+i*sin(±2π/3) =-1/2±i√3/2. In other words, from the initial line, 6th power is 1 rotation, cubed is half rotation, and squared is 2π/3(120 °) rotation. Therefore, It can be solved as x^65={(x^6)^10}*(x^3)*(x^2)={(1)^10}*(-1)*(x^2)=(1)*(-1)* (x^2)=-(x^2), therefore, x^65+1/(x^65)=-(x^2)-1/(x^2) =-{( x ^2)+1/(x^2)}=-{(-1/2+i√3/2)+(-1/2-i√3/2)} or -{(-1/2)-i√3/2)+(-1/2+i√3/2)}=-(-1/2-1/2) =-(-1) = 1.
No matter how much the multiplier of x increases, if you can understand that the value "turns on the unit circle", then all you have to do is know which position [1+i*0, 1/2+i*(√3/2),-1/2+i*(√3/2), -1+i*0, -1/2-i *(√3/2), 1/2-i*(√3/2)] to go by the multiplier.
Pretty good!
My solution:
Let An = x^n + 1 / x^n
A1 = 1
We are trying to find A65
(x^n-1 + 1 / x^n-1) * (x + 1/x) = x^n + 1 / x^n + x^n-2 + 1/x^n-2
An-1 * A1 = An + An-2
An-1 = An + An-2
An = An-1 - An-2 defines a recursive sequence
Write the sequence:
A0 = 2
A1 = 1
A2 = -2
A3 = -2
A4 = -1
A5 = 1
------------- repeats every 6 values
A6 = 2
A7 = 1
...
65 mod 6 = 5, so A65 = A5 = 1
Nice!
Salute to Prof. You have attempted for possible ways for solution. Good presentation. worthy tips to student of Mathematics.
Thanks a lot
Substitute x=-w where w is complex cube root of unity
Then x^65=-w^65=-w^2
x^-65=-w^-65=-w^130=-w
-w^2-w=1
Hence f(1)=1
You are genius ,dear Advay
Very nice!
What is the domain of f? Does it have to be the set of all complex numbers? Can f be a real value function for this problem to make any sense?
What was the formula you used in 2nd method when you raised the trigonometric sum to the power of 65?
De Moivre's formula
Excellent! Wonderful solution! Thank you sir!
Np. Thank you!!! 🧡
How can you assume x^2 +1 = x when they are actually not equal at all ?!!
Even the 1st method uses complex numbers. If we restrict x to be a real number, then x+1/x can never be 1.
Faster method 3:
x³ = -1
x⁶ = 1
x⁶⁵ = x⁶⁶/x = 1/x
1/x⁶⁵ = x
x⁶⁵ + 1/x⁶⁵ = 1/x + x = x + 1/x = 1
Although the 3rd method is the fastest way, I love the 2nd method because it is more like I am doing the mathematics. 😁😁😁
x+1/x is continuous on the set of real numbers excluding 0..consider the sequence Xn(say) (1+1/n) converging to 1 so f(Xn) converges to f(1).. So limit(n tends to ♾️) f(Xn)=lim(n tends to ♾️)(1^65+(1/n)^65)=1=f(1).. So f(1)=1..
I was about to solve it using your second method until I found out I had to raise that huge expression to the 65th power!
😁
cosx + isinx = e (ix)
and e(ix)^65 = e(65ix)
so very simple
I used limit (I don't know if it is ok to do this):
as x → ∞, x+1/x → 1 and x⁶⁵ + 1/x⁶⁵ → 1
⇒ f(1) = lim x→∞ [f(x+1/x)] = lim x→∞ [x⁶⁵ + 1/x⁶] = 1
how does x + 1/x go to 1 as x goes to infinity?? you should think this over again
@@shacharh5470 yea, my bad
i calculated x+1/x = (x+1)/x
Very good question sir
Thank you so much
You are genius , dear professor
Oh, thank you for the kind words! 🥰
x+1/x=1 percio'.x^2=x-1......calcolo facilmente x^4,x^8,x^16,x^32,x^64.....x^65+1/x^65=1
Congratulations !!!🎉🎊 👏🏾👏🏾👏🏾👏🏾
Thank you! 🧡
All 3 methods are good thanks for your vidéos 👍
Np. Thank you!!!
Cube roots of negative 1
How can we divide 1-x by 1-x when we are evaluating at x=1?
We’re not, we’re evaluating it at an x that satisfies x+1/x=1, which x=1 clearly does not.
@@ConManAU Thanks for the response!
wow what a mind-blowing solution
Glad to hear that!
I solved for the function itself and I got f(x) = (x+-sqrt((-x)^2-4)/2)^65 + (2/(x+-sqrt((-x)^2-4))^65
Yes, but it's only valid for x>2 or x
Wonderful!
Thanks!
"Bengt Larsson" is a rather common name. At least in Sweden and in the over-40 demographics.
Interesting!
Good. Both methods
.!
Many many thanks
Hi. Marvelous video)
I want to ask you about problems, where did you find such interesting themes and examples?
You explored it in the internet or there are cool books? Pls recommend😊
Thank you!
Books, internet, friends and myself...
This one I made up (something that I enjoy doing) but that's easy to do if you've dealt with these kinds of equations before.
Very very nice methods
Thanks!
Very interesting
And what's x^64 + 1/x^64 ?
-1
f(x+1/x)=x^65+1/x^65 find f(x)
You can see the full solution in my comment above.
good question
Awesome
I love it
Thank you! 🧡
Nice
👍
-1
2
may be but not sure.
substitute x=1
f(2)=2
f(1)=1
2