🧡🧡🧡 Welcome to the official RU-vid channel of SyberMath. I love solving Algebra, Calculus, and Number Theory problems that are fun and challenging. Even though I taught math for a while, I do not consider myself a mathematician. I'm someone who simply loves to solve math problems for fun because math is 🧡. If you're looking for fun and somewhat challenging problems but not necessarily looking for rigor and serious math and/or a Math Olympiads, Math Competitions, SAT, and JEE aspirant, then you've come to the right place!!! 🤩 I tend to introduce more than one method to solve a problem if possible. I also include graphs made with Desmos whenever appropriate and sometimes show solutions from Wolfram Alpha. I'm always amazed by my audience's proficiency in math and intellectual perspicacity. I also have a secondary channel for shorts and lecture videos: youtube.com/@SyberMathShorts Follow Me on Twitter: twitter.com/SyberMath Happy Solving and Happy Watching!!! 🧡🧡🧡
p+q+r=29 pq+qr+rp=244 pqr=576=2²3²4² The 3rd eq'n gives a clue. (p,q,r)=(4,9,16) or its permutation [sqrt(p),sqrt(q),sqrt(r)]=(2,3,4) or its permutatiom
You leave Sybermath alone for a day, and when you come back he has invented these horrors of horrors! I suppose substitution will somehow do the trick.
Feel like a mouse being put into a maze. The question is: how the experience of finding a way out in this maze could help a mouse finding the way out in next one?
Synthetic Division fails for x + 1 = 0 into 1 0 -3 -2 giving a false (x + 1) (x^2 - x - 2) result. However, choosing the non repeating x - 2 = 0 by going into 1 0 -3 -2 we get x^2 + 2x + 1 = 0 part by Synthetic Division. Synthetic Division works if we choose the correct root that divides into the Polynomial. We then have, easier by Synthetic Division of (x - 2) into 1 0 -3 -2 forms (x - 2)(x^2 + 2x + 1) = 0 correct answer! 😂🤣
Actually x^2 - x - 2 gives the (x + 1)(x - 2) Quadratic Equation factoring. I got myself confused erroneously thinking I would get the repeated x + 1 or x^2 + 2x + 1 result. This is when using the easier method Synthetic Division causes confusion that isn't contradictory. I am just guessing why Synthetic Division is not a lectured method 3 we get in high school. 🤯
Another way of showing that P(x) is linear is assuming a is a root of P (maybe complex) So P(a)=0 then P(P(a)+1)=9a+7 P(1)=9a+7 Now since P is a function, P(1) has a unique value, so 9a+7 has a unique value, so a has a unique value. i.e, P has a unique root, hence it is linear. Notably, this can be used to prove uniqueness of the root even if P is not a polynomial, so it may help in generalising the functional equation edit: one slight correction, for a general function the root may or may not exist. i.e no. of roots are either 0 or 1
Someone isn't posting his detailed analysis anymore so I will do the analysis. Find all functions y : 𝔻→ℝ such that for all x in 𝔻 ; y'(x) = xy(x) / (x^2 + y(x)^2) where 𝔻 is a subset of ℝ . First note the restrictions caused by the equation: 1. x^2 + y(x)^2 is in the denominator so it can't equal to 0 but it is the sum of two nonnegative terms so it can only equal 0 if both terms are 0. meaning that for x = 0 , y(x) can't be 0 We make the substitution y(x) = xu(x) (I think we can do that because the equation is homogeneous) but we have to be careful! The substitution doesn't work for x = 0 because y(0) would be 0, so one must check the case of x = 0 seperately. If x = 0 then y'(0) = 0. We will come to that later. As for x ≠ 0, after the substitution we obtain u(x) + xu'(x) = (x^2)u(x) / (x^2 + x^2u(x)^2) ie. u(x) + xu'(x) = u(x) / (1 + u(x)^2) ie. xu'(x) = u(x) / (1 + u(x)^2) − u(x) ie. xu'(x) = −u(x)^3 / (1 + u(x)^2) Now we divide both sides by x(−u(x)^3) / (1 + u(x)^2 which assumes that x and u(x)^3 are nonzero. For x, this is already true. However, for u(x) we don't know. We will call S the set of values that set u(x) to zero. After division we get that for all x in 𝔻\S ; u'(x)(1/u(x)^3 + 1/u(x)) = −1/x We will then recall the formulas of derivatives and the properties of differentiation to find u. However, We will need to solve the equation in two cases according to the sign of x. For x > 0, and because we are solving the equation over an interval, and since u is continuous, it doesn't change its sign, we will have 2 cases. When u(x) is positive, the equation becomes 1/(−2u(x)^2) + ln(u(x)) = −ln(x) + a ie. −1/2ln[exp(1/u(x)^2)] −(1/2)ln[1/u(x)^2] = −ln(x) + a ie. −(1/2)ln[exp(1/u(x)^2)/u(x)^2] = −ln(x) + a ie. ln[exp(1/u(x)^2)/u(x)^2] = 2ln(x) −2a = ln(x^2) ie. exp(1/u(x)^2)/u(x)^2 = Ax^2 where A > 0 ie. 1/u(x)^2 = w(Ax^2) ie. u(x) = 1/√W(Ax²) (we can invert w(Ax^2) because for all x > 0, this is always nonzero) A similar calculation for the case when u(x) is negative leads to u(x) = −1/√W(Ax²) (just replace u(x) by −u(x) everywhere) For x < 0, we will get the same results. Now what's left is finding S. As always, S could be the entire domain making u(x) = 0 which is verified by the equation by that's te trivial case. We will consider now that the values arround the point s in S do not all set u(x) to 0 meaning that u(x) obeys the expression we found earlier. We know that u(x) is continuous so lim (x→s) u(x) = u(s) ie. lim (x→s) −1/√W(Ax²) = 0 but there is no s such that this limit eqaution is verified so S is empty and there is no piecewise solutions. Now we come back to the substitution and then compute lim (x→0) y(x) to see whether y should be defined at 0 which is something I don't know how to do. The final solution is then any function y : ℝ\{0}→ℝ such that for all x in ℝ+, y(x) = x/√W(Ax²) or y(x) = −x/√W(Ax²) and for all x in ℝ−, y(x) = x/√W(Bx²) or y(x) = −x/√W(Bx²) where A and B are both > 0. If y happened to be defined at 0, the constants A and B would still be independent.
I realized I overlooked something: lim (x→0, x>0) y(x) and lim (x→0, x<0) would be dependent on A and B respectively. If they happened to be the same, then only in that case A and B are independent. If not, then A and B should be equal. And of course, if the limit happened to be 0, then by continuity y(0) must be 0 which is something established to be impossible in the beginning.
This doesnt work on every function. When stating a problem, you should give the constraints. A function is a mapping between 2 sets. There is no need for the mapping to follow any pattern, it can be absolutely valid yet still a valid function. I am not sure what the constraints here are, but surely, the function must be differntiable, not made of several functions stiched togethet, contiguous and have a relationship between any f(x) and f(x+n) which is a primitive function. This seems dishonest to me. The simplest proof is a function defined at only 1 point. Maybe even a constant function.
I got an odd looking solution by using angles. Using polar coordinates, and substituting everything, I got the complex answer y = ix / sqrt(2), and it does actually work in the original equation
Buenos días Señores SyberMath, reciban un cordial saludo y un agradecimiento por esta obra de arte. Por favor un refuerzo en la determinación de las raíces complejas con los logaritmos, gracias por la atención que se me brinde. Éxitos.
problem dy/dx = xy/(x²+y²) let p = x/y y = x/p dy/dx = (p - x dp/dx) /(p²) and our differential equation becomes (p-x dp/dx)/(p²) = p/(p²+1) solving for dp/dx, dp/dx = [p-p³/(p²+1)]/x which separates nicely into dp/[p-p³/(p²+1)] = dx/x dp/[(p³+p-p³)/(p²+1)] = dx/x dp/[p/(p²+1)] = dx/x (p²+1)dp/p = dx/x (p+1/p)dp = dx/x ∫(p+1/p)dp = ∫dx/x p²/2+ln p = ln x+c p = x/y ln x +c = (x/y)²/2+ln x/y ln x +c = [x²/(y²)]/2 + ln x - ln y [x²/(y²)]/2 - ln y = c x²/(y²) - 2 ln y = c, c is still a constant x²/(y²) + -2 ln y = c x² y⁻² + ln y⁻² = c Exponentiate both sides. y⁻² e^(x² y⁻²) = eᶜ Multiply both sides by x². (x²y⁻²) e^(x²y⁻²) = eᶜx² The left side is clearly W(eᶜx²), a Lambert's W function application. x²y⁻² = W(eᶜx²) y² = x²/W(eᶜx²) y = ± x/√W(eᶜx²) answer y ϵ { -x/√W(eᶜx²), x/√W(eᶜx²) }, where c is a constant
I found y = ±√[e^W(eᶜx²)/eᶜ], which looks different from your result y = ± x/√W(eᶜx²), but is equivalent. Proof: √[e^W(eᶜx²)/eᶜ] = x/√W(eᶜx²) implies e^W(eᶜx²)/eᶜ = x²/W(eᶜx²). Cross multiplying, W(eᶜx²) e^W(eᶜx²) = eᶜx². Applying W to both sides, we have W(eᶜx²) = W(eᶜx²).
@@DonEnsley-yi2ql If y happens to be undefined at 0, then y could be piecewisely defined with different constants for x>0 and x<0 because the domain is composed of 2 intervals. Also, y could have different signs for each interval. For example, for the positive interval use the positive expression and for the negative interval use the negative expression. There would be 4 solutions.
