I solved it in much simpler way: square both sides-> x*sqrt(x^2sqrt(x^3...))=9 , then "take out" an X^2 from each radical, you get x*x*sqrt(xsqrt(x^2sqrt(...)))=9 so, the "sqrt block" can be replaced with "3", which is the starting equation, and then you easily get 3*X^2=9 hence the final solution...
@theamazingfrogman11 ай бұрын
Beautiful problem and solution
@SyberMath11 ай бұрын
Thank you! 😍
@Physics_HB11 ай бұрын
Thanks❤
@SyberMath11 ай бұрын
My pleasure ❤️
@emanuellandeholm56576 ай бұрын
This power series is essentially the derivative of a power series that you can do geometric summation of.
@kianmath7111 ай бұрын
Great video😊
@SyberMath11 ай бұрын
Thank you 😁
@nasrullahhusnan22896 ай бұрын
The problem can be written as 3=x^(½+¼+⅛+...) --> 3=x
@JourneyThroughMath11 ай бұрын
That method for solving the infinite sum is handy...i shamefully used wolfram alpha
@goldfing589811 ай бұрын
I assume that under the second root, it is x^2 * sqrt(x^3), not x * second root of x^3. You should put more space between x^2 and the root sign. So all roots are square roots, I assume.
@SyberMath11 ай бұрын
yes. First of all 2 as an index is never written. Square root is understood if there is no index. Second, the index is usually over the part of the radical sign that looks like a V, much closer to the radical than it's to the expression to the left of it. However, I agree with you, spacing out will make it more clear
@goldfing589811 ай бұрын
@@SyberMath I'm a math teacher and when it comes to root computations, I omit the root index 2 when there are only square roots (e.g. for almost any root equation in school books). But when there are roots of several types (e.g. simplify a root expression with square roots and cubic roots etc.), I intentionally write down the 2 for square roots, for two reasons: 1.) To assert to me and the reader that I didn't forget any root index. 2.) To easen the computation. For example, I often use the nice rule that in the case of directly nested roots, it is allowed to swap the roots, or to merge them into one root by multiplying their root indices.
@goldfing589811 ай бұрын
One remarkable situation was when I had to prove the trigonometric formulae for the "casus irreducibilis" of the general cubic equation ax^3 + bx^2 + cx + d = 0. After applying the linear Tschirnhaus transformation. x = y - b/(3a), the depressed cubic was something like y^3 + 3py + 2q = 0 and its quadratic resolvent something like z^2 + 2qz - p^3 = 0. Its solutions were z1,z2 = -q +- sqrt(q^2 + p^3). The Cardano-Tartaglia formula demands that y = u + v with u^3 = z1 and v^3 = z2. So I had to compute the cubic roots of z1 and z2. In the case of a negative discriminant D = q^2 + p^3 < 0 which can only happen if p is negative, we have the problem of computing the cubic roots of the two complex conjugate numbers z1 and z2. In this context, I did the following simplifcation of the absolute value of the cubic roots: cubrt(sqrt(-p^3)) = sqrt(cubrt(-p^3)) = sqrt(-p) which can be computed in real arithmetics, because p is negative, thus -p is positive 🙂
@goldfing589811 ай бұрын
The z1,z2 formula got somewhat corrupted. It meant z1, z2 = minus q plus/minus square root(q^2 + p^3).
@georgesbv111 ай бұрын
X cannot be negative since it's an infininite product of non integer powers. Positive guarantees convergence
@yusufdenli936311 ай бұрын
Very nice 😉
@SyberMath11 ай бұрын
Thank you
@JourneyThroughMath11 ай бұрын
Well, it would be an infinite sum of n/2^n which equals 2. X^2=3 so x =+-root3
@farhansadik542311 ай бұрын
Thankfully, after watching you videos for so long, i remembered a similar problem, that you had done before, I also remembered a formula that I derived after becoming inspired! So for √x√x^3√x^5√x^7...= x^3 This is also very fun to do, though it's a bit trickier! You should try this too!
@SyberMath11 ай бұрын
Thanks for sharing 😍
@DarsheelAE11 ай бұрын
i got x= 3root(3root(3))
11 ай бұрын
x.x.3=9. x = ✓3
@comdo77711 ай бұрын
asnwer=2 isit
@rakenzarnsworld211 ай бұрын
x = 3
@E.h.a.b11 ай бұрын
√x √x^2 √x^3 √x^4 √x^5 ........ (x^1)^1/2 = x^(1/2) 1/2 √√x^4 √x^3 √x^4 √x^5 ........ (x^4)^1/4 = x^(4/4) 1/2 + 2/4 √√√x^11 √x^4 √x^5 ........ (x^11)^1/8 = x^(11/8) 1/2 + 2/4 + 3/8 √√√√x^26 √x^5 ........ (x^26)^1/16 = x^(26/16) 1/2 + 2/4 + 3/8 + 4/16 √√√√√x^57 ........ (x^57)^1/32 = x^(57/32) 1/2 + 2/4 + 3/8 + 4/16 + 5/32 Assume we will finally get X^S where ∞ ∞ S = ∑(n/2^n) = ∑(n * (1/2)^n) ---------> [1] n=1 n=1 I will use the following formula to get S ∞ K ∑K^n = ------ where |K| < 1 ---------> [2] n=1 1-K Differentiate both sides we get ∞ 1 ∑n K^(n-1) = ----------- n=1 (1-K)^2 Multiply both sides by ( K ) we get ∞ K ∑n K^n = ----------- ---------------------> [3] n=1 (1-K)^2 To get the value of S we put K = 1/2 in [3] S = (1/2)/(1-1/2)^2 = 1/(1/2) = 2 x^2 = 3 x = √3 Answer