BMAD 64 Report post Posted March 25, 2013 There are three solutions. Can you find a formal proof for all three? Quote Share this post Link to post Share on other sites
0 bonanova 84 Report post Posted March 25, 2013 x=2 and 4. Proof: 2^{2} = 2^{2} 4^{2} = (2^{2})^{2} = 2^{4}. The third solution is x = -.76666468...., found numerically, looking for closed form. Quote Share this post Link to post Share on other sites
0 bonanova 84 Report post Posted March 25, 2013 Using natural logarithms, the equation becomesx = 2 ln(x) x = 2 = 2 ln (2) = 2 x 1 = 2 x = 4 = 2 ln (4) = 2 x 2 = 4 The third root is negative and solvesx = -2 ln (x) Which has no algebraic solution. Numerically, it'sx = -0.766664695 .... Interested to find out what you mean by formal proof. Quote Share this post Link to post Share on other sites
0 dark_magician_92 4 Report post Posted March 25, 2013 roots are easy to find and 2 roots may be somehow proved. 3rd is difficult to prove. Using log base e, 2 lnx = x ln 2 x/lnx = 2/ln2, By comparison, x=2 and also LHS= 4/2lnx, again compare, x=4. Quote Share this post Link to post Share on other sites
0 Federico 0 Report post Posted May 7, 2019 Starting from the raising power operator, defined between two numbers x and y over a number field, x^{y }is not the same as y^{x}, due to the missing commutativity property respect than the other sum and multiplication rank binary operators. So this post is questioning for wich value of x in natural, integers, reals (or complex) number fields this identity is verified: x^{2} = 2^{x} . This is what we call an equation to solve. This problem could also be interpreted geometrically asking in which points the parabola y=x^{2} intersect the esponential function y=2^{x}. Before we start exploring over the natural number (positive integers), we know that 2^{x} > 0 and x^{2 }≥ 0 for every x Reals. So, we can exclude immediatly the case for the integer x=0 wich implies: 0^{2 }< 2^{0,}, that is 0 < 1. Furthermore, for the same reason and due to negative power of 2, for every negative integer x ≤ 1, we have 2^{x} < x^{2} . So, for positive integers x>0, both 2^{x}and x^{2 }are crescent functions : 2^{x+1 }≥ 2^{x} and (x+1)^{2 }≥ x^{2}, and by the fact that: for x=1, 1^{2} <2^{1} (that is 1 < 2), and for x=3, 3^{2 }>2^{3 (}that is 9>8), this implies that an integer solution is between 1 and 3. Hence, our 1st trivial solution is for x=2: 2^{2} = 2^{2 }= 4 . In a similar way, from the fact that at x=5 , we have 2^{5} = 32 > 5^{2} = 25, we candidate next integer solution to be between 3 and 5, that is x=4. Analizing the number 4 as result of this 1st trivial solution, It is interesting to observe that the number 2 let the 4 = 2+2 = 2*2 = 2^2 = ... , for every rank of the binary operator used. 4^2 = (2^2)^2 = 2^(2*2) = 2^(2+2) = 2^4 = 16. (Notice that just for the positive integer 2 you can make (2^2)^2 = 2^(2^2) ) From this point on, infact, 2^x will exceed x^2 for every real x>=4, so no more positive real solution after 4. Infact for every integer x>0 (x+4)^{2} ≤ 2^{(x+4)} , applying log in base 2 to both members, we can compare 2 log_{2}(x+4) ≤ x+4 ≤ x+4 + x+4 ≤ 2(x+4) log_{2}(x+4) ≤ x+4 , for every real x>0. So for every x>4 we proved that x^{2} ≤ 2^{x}. But it's not the only interval where 2^{x} exceeds x^{2}. infact, x^{2 }= 2^{x} , can be rewritten as 2nd grade equation giving the: x = 2^{x} , and x = 2^{-x} solutions. Using the fact that (-x)^{2 }= 2^{x} , or (x)^{2 }= 2^{-}^{x} x = 2^{-x/2} From the theorem of the fixed point applied to our exponential, starting from x_{0}=1, and proceeding recursively x_{1} = -2^{x0/2 }= -√2 x_{2} = -2^{x1/2} = -√2^{-1/}^{√2} x_{3} = -2^{x2/2} = -√2^{-}^{√2/2} ... x_{n} = -2^{xn-1 } = -√2^{-√2^(-√2^....n-1Times....^-√2))} The greater will be n and the smaller will be the difference x_{n+1 }-x_{n.} The limit for n that tends to infinity of x_{n }is equal to value x: x_{1}_{ }= 1 x_{1}_{ }= -1.41421356237310... x_{2 }= -0.612547326536066... x_{3 }= -0.808727927081827... x_{4 }= -0.755569324854238... x_{5 }= -0.769618475247432... x_{6 }= -0.765880261444542... x_{7 }= -0.766873153047868... x_{8 }= -0.766609309719418... x_{9 }= -0.766679412545762... x_{10 }= -0.766660785691404... x_{11 }= -0.766665734944187... x_{12 }= -0.766664419898679... x_{13 }= -0.766664769313767... x_{14 }= -0.766664676472171... x_{15 }= -0.766664701140715... x_{16 }= -0.766664694586141... x_{17 }= -0.766664696327729... x_{18 }= -0.766664695864979... x_{19 }= -0.766664695987935... x_{20 }= -0.766664695955265... x_{21 }= -0.766664695963945... x_{22 }= -0.766664695961639... x_{23 }= -0.766664695962252... x_{24 }= -0.766664695962089... x_{25 }= -0.-66664695962132... x_{26 }= -0.766664695962121... x_{27 }= -0.766664695962124... x_{28 }= -0.766664695962123... x_{29 }= -0.766664695962123... The value asimptotically converge quickly according the desired precision. Another method to find a closed form using the Lambert function W(z) = z e^{z}, which gives the solution in function of F. Prowedi Quote Share this post Link to post Share on other sites
There are three solutions. Can you find a formal proof for all three?
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