Is This A New Formula For Infinitely Nested Radicals?

$x+a=\sqrt[3]{a(x^2+3ax+a^2)+x^2\sqrt[3]{2a(x^2+3(2a)x+(2a)^2)+x^2\sqrt[3]{4a(x^2+3(4a)+(4a)^2)+x^2\dotsm}}}$ $Derivation:$I made use of the following identity; $x+a=\sqrt[3]{(x+a)^3}\\=\sqrt[3]{x^3+3ax^2+3a^2x+a^3}\\=\sqrt[3]{x^3+2ax^2+ax^2+3a^2x+a^3}\\=\sqrt[3]{ax^2+3a^2x+a^3+x^3+2ax^2}\\=\sqrt[3]{a(x^2+3ax+a^2)+x^2(x+2a)}$ In the 2nd step I wrote $3ax^2$ as $2ax^2+ax^2$.In the 3rd step I rearranged the terms.In the 4th step I factored out the $a$ from the first 3 terms and I factored out the $x^2$ from the last 2 terms.Manipulating $x+2a=\sqrt[3]{x^3+3(2a)x^2+3(2a)^2x+(2a)^3}$ in the same way,I got $x+2a=\sqrt[3]{2a(x^2+3(2a)x+(2a)^2)+x^2(x+4a)}$.In the same way I got the expression for $x+4a,x+8a,x+16a\dotsm$.Now I observed that each of these can be substituted in the previous expression.Substituting my way upto $x+a$,I got the expression written at the beginning.Now,I want the people on Brilliant to verify this expression and to see whether it works.Please help me!!!

Note by Abdur Rehman Zahid
5 years, 3 months ago

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Put a = 0

R.H.S becomes 0 but L.H.S is still 'x'

So I don't think it's correct

- 5 years, 3 months ago

Because you are dealing with infinite sums/roots, you have to be really careful of what the proper interpretation is.

Of course, it could be that the statement is not true for $a = 0$, but is true for all other real numbers.

Staff - 5 years, 3 months ago

Just looking back at this note and thinking......That was a long time ago

- 4 years, 2 months ago

Please explain how R.H.S becomes zero

- 5 years, 3 months ago

Every term on R.H.S is multiple of 'a' you can see itself

- 5 years, 3 months ago

If we rearrange the formula for $x+a=\sqrt[3]{x^3+3ax^2+3a^2x+a^3}$, you get $x+a=\sqrt[3]{a(x^2+3ax+a^2)+x^2(x+2a)}$ In this expression if you replace a by 2a, you get $x+2a=\sqrt[3]{2a(x^2+3(2a)x+(2a)^2)+x^2(x+4a)}$Replacing this expression in the expression for x +a, you get $x+a=\sqrt[3]{a(x^2+3ax+a^2)+x^2\sqrt[3]{2a(x^2+3(2a)x+(2a)^2)+x^2(x+4a))}}$In this way if we continue deriving the expression for x+4a,x+8a,x+16a etc and replacing them in the original formula for x+a we will get the original formula .So R.H.S does not become zero you just have to replace the value of x+2a instead of the next radical and then set a=0 and after simplifying you get $x=\sqrt[3]{x^3}$

- 5 years, 3 months ago

Putting $a \,=\, 0$ yields another infinitely nested series of $x$, which on simplification is equal to $x$.

- 3 years, 10 months ago

Yup

- 3 years, 10 months ago

if a=0 then x also equal to 0 as x-x=0

- 5 years, 1 month ago

Ah yes. Isn't it amazing how much can be achieved in a year?

Staff - 4 years, 2 months ago

Yeah.I didn't even know a thing back then.Brilliant has helped me a lot

- 4 years, 1 month ago