Is This A New Formula For Infinitely Nested Radicals?

x+a=a(x2+3ax+a2)+x22a(x2+3(2a)x+(2a)2)+x24a(x2+3(4a)+(4a)2)+x2333x+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:Derivation:I made use of the following identity; x+a=(x+a)33=x3+3ax2+3a2x+a33=x3+2ax2+ax2+3a2x+a33=ax2+3a2x+a3+x3+2ax23=a(x2+3ax+a2)+x2(x+2a)3x+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 3ax23ax^2 as 2ax2+ax22ax^2+ax^2.In the 3rd step I rearranged the terms.In the 4th step I factored out the aa from the first 3 terms and I factored out the x2x^2 from the last 2 terms.Manipulating x+2a=x3+3(2a)x2+3(2a)2x+(2a)33x+2a=\sqrt[3]{x^3+3(2a)x^2+3(2a)^2x+(2a)^3} in the same way,I got x+2a=2a(x2+3(2a)x+(2a)2)+x2(x+4a)3x+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+16ax+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+ax+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
4 years, 11 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

Krishna Sharma - 4 years, 11 months ago

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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 a = 0 , but is true for all other real numbers.

Calvin Lin Staff - 4 years, 11 months ago

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

Abdur Rehman Zahid - 3 years, 9 months ago

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Please explain how R.H.S becomes zero

Abdur Rehman Zahid - 4 years, 11 months ago

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

Krishna Sharma - 4 years, 11 months ago

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@Krishna Sharma If we rearrange the formula for x+a=x3+3ax2+3a2x+a33x+a=\sqrt[3]{x^3+3ax^2+3a^2x+a^3}, you get x+a=a(x2+3ax+a2)+x2(x+2a)3x+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=2a(x2+3(2a)x+(2a)2)+x2(x+4a)3x+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=a(x2+3ax+a2)+x22a(x2+3(2a)x+(2a)2)+x2(x+4a))33x+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=x33x=\sqrt[3]{x^3}

Abdur Rehman Zahid - 4 years, 11 months ago

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

Aditya Sky - 3 years, 6 months ago

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Yup

Abdur Rehman Zahid - 3 years, 6 months ago

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

Muhammad Yousaf Khan - 4 years, 9 months ago

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

Calvin Lin Staff - 3 years, 9 months ago

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

Abdur Rehman Zahid - 3 years, 9 months ago

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