If $f(x)=\sqrt{1+\sqrt{x+\sqrt{x^2 +\sqrt{x^3 +\cdots}}}},$ then find the value of $f(4).$

Question

My attempt: $f(x) = \sqrt{1+{\sqrt{x}} f(x)} \implies f(4) =\sqrt{1+2f(4)} \implies f(4)(f(4)-2)=1.$

I don't know how to proceed from here. The answer should be $2.$ Where am I getting wrong?

Kindly, explain it.Thank you for any guidance!

Check your third step from $f(4) = \sqrt{1 + 2f(4)}$ to $(f(4) - 3)(f(4) + 1)$ closely again :) — I am a person, May 18 '22 at 02:03
The first asserted equality $f(x)=\sqrt{1+\sqrt x f(x)}$ is incorrect. — Greg Martin, May 18 '22 at 02:33
Why do you believe $f(x) = \sqrt{1+\sqrt{x} f(x)}$? $xy=1$ does not imply $x=1$ or $y=1$. That only works with zero: $xy=0$ implies $x=0$ or $y=0$. — aschepler, May 18 '22 at 02:33
@Greg Martin, Would you explain why $f(x)=\sqrt{1+\sqrt x f(x)}$ is incorrect? — Debrogli, May 18 '22 at 02:37
@Debrogli Multiply $\sqrt{x}f(x)$ and go out to about three or four nested radicals. It doesn't give the right sequence of exponents. — Brian Moehring, May 18 '22 at 02:53
Does this help? https://math.stackexchange.com/q/1785454/1012971 — , May 18 '22 at 03:46

score 1 · Answer 1 · edited May 18 '22 at 07:08

1

For $y>0$, \begin{eqnarray} 1+y & = & \sqrt{y^2+(2y+1)}\\ & = & \sqrt{y^2+\sqrt{(2y+1)^2}}\\ & = & \sqrt{y^2+\sqrt{4y^2+(4y+1)}}\\ & = & \sqrt{y^2+\sqrt{4y^2+\sqrt{(4y+1)^2}}}\\ & = & \sqrt{y^2+\sqrt{4y^2+\sqrt{16y^2+(8y+1)}}}\\ & \vdots & \\ & = & \sqrt{y^2+\sqrt{4y^2+\sqrt{4^2y^2+\sqrt{4^3y^2+\sqrt{4^4y^2+\cdots}}}}} \end{eqnarray}

If you set $y=1$ you get the value of your function for the special case $x=4$, i.e. $f(4)=1+1=2$.

edited May 18 '22 at 07:08

Greg Martin

92,241

answered May 18 '22 at 04:09

Peder

2,232

Note that the relations only hold for $y>0$ as the left hand is 1 for $y=0$ while the limit appears to be 0. – Peder May 18 '22 at 04:12

If $f(x)=\sqrt{1+\sqrt{x+\sqrt{x^2 +\sqrt{x^3 +\cdots}}}},$ then find the value of $f(4).$

1 Answers1