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6 solutions
I haven't learnt calculus yet.... : ( : (
Can you please explain to me the step of adding exponents ? @Caleb Townsend
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Yes, since we're multiplying 5 1 / 2 × 5 1 / 4 × . . . × 5 1 / 2 x , the resulting exponent should be 1 / 2 + 1 / 4 + 1 / 8 + . . . + 1 / 2 x If you simplify the expression, it becomes something like 4 / 8 + 2 / 8 + 1 / 8 , or 8 / 1 6 + 4 / 1 6 + 2 / 1 6 + 1 / 1 6 , etc. When you simplify this, the result will be 2 x 2 x − 1 .
More conceptually, if you were to forgo the final term of that sum, and replace it with the previous term, the sum would be equal to 1 ; and since the previous term is equal to twice the final term, you can algebraically find the sum using substitution.
Do not understand calculus properly to use it in solving process !
Let 5 5 5 ⋯ = x
5 x = x
Squaring both sides ,
⇒ 5 x = x 2
⇒ x = 5
Notice that, x = 0 also satisfies the equation 5 x = x 2 . You should describe why x = 0 is not a valid solution.
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I don't think x = 0 is a valid solution of the problem, it is only a solution for the equation proposed by Nihar. It is more visible if you rewrite 5 5 5 ⋯ as 5 1 / 2 × 5 1 / 4 × 5 1 / 8 × ⋯ then you see that it can never be zero.
Yeah! I never thought of that! Thanks. If you have a complete solution , it would be beneficial for all of us if you post it!
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Sorry for the late reply. Sir Caleb Townsend has already showed a complete solution using calculus. If you don't know calculus, it can also be solved using inequality. (that f ( x ) > 0 )
Indeed this solution is incomplete. It's easy to jump the gun in math
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If you have a complete solution , I request you to post it.
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@Nihar Mahajan – In addendum to your solution, you could note that no matter how many iterations of the square root are taken, the result will not approach 0 . Therefore out of the two possible solutions x = 5 and x = 0 , the only valid solution is x = 5 .
@Nihar Mahajan – I think x > 0 would be sufficient to complete the solution. Although another way to look at it would be to write the nested radical as: 5 2 1 × 5 4 1 × 5 8 1 × . . . = 5 a w h e r e a = 2 1 + 4 1 + 8 1 + . . . Now we can think of the infinite sum by cutting a ribbon in half, then cutting one the halves in half again and so on. Of course we only have one ribbon so we produce an answer of 5.
Can you say why x can't be 0 ?
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In this problem, theoretically x can be 0, as the equation resolves finally to x^2 - 5x=0, i.e x(x-5) =0
But coming back to original assumption that x = root 5 continued, we know x has to be value greater than 2, as root 5 will be greater than root 4, ie 2. Hence x= 0 can not be a solution, as it defies the original assumption. Therefore, what remains is x=5, which is the only solution.
It should be x^2 -x-5=0 right?
i like your simple calculation
This is the way i did :D
This is how I got it - like that of Fauzi Yul Chaidir,
OK
suppose :
x = 5 5 5 …
so that:
x = 5 x ,
square the x
x 2 = 5 x → x x 2 = 5 → x = 5
this cant be done when x=0 is a solution
Y= 5^1/2 + 5^1/4 + 5^1/8+.... Log y = log 5 ( 1+ 1/2 + 1/4+1/8+....) Logy = log 5 . y = 5.
You can make a geometric progression with the exponents. If you take'em and sum using the formula for the sum of the terms of a convergent gp, the result will be 5 to the 1st power, that is 5
Consider a function f ( x ) = 5 5 5 . . . 5 x so that f ( 1 ) = 5 , f ( 2 ) = 5 5 , f ( 3 ) = 5 5 5 , and so forth. Note that f ( x ) is precisely f ( x ) = 5 1 / 2 × 5 1 / 4 × . . . × 5 1 / 2 x Adding the exponents gives f ( x ) = 5 ( 2 x − 1 ) / ( 2 x ) Now taking the limit of f ( x ) , x → ∞ lim f ( x ) = 5 lim x → ∞ ( 2 x − 1 ) / ( 2 x ) since 5 x is continuous for x . Applying L'Hopital's rule, x → ∞ lim ( 2 x − 1 ) / ( 2 x ) = x → ∞ lim ( 2 x ln 2 ) / ( 2 x ln 2 ) = 1 therefore x → ∞ lim f ( x ) = x → ∞ lim 5 1 = 5