Find the smallest possible value of the constant A such that ln(x) 0.

Question

Find the smallest possible value of the constant A such that ln(x) <= Ax^2 for all x>0.

ganeshie8 · Answer

Hey

yb1996 · Answer

Hello

ganeshie8 · Answer

tried anything ?

yb1996 · Answer

No, I'm not really sure what I'm supposed to do to be able to get the answer

ganeshie8 · Answer

Have you understood the question ?

ganeshie8 · Answer

Notice that `lnx` grows slowly compared to `x`

ganeshie8 · Answer

Honestly i also don't know how to solve this problem, im still thinking..

yb1996 · Answer

Yes, so I'm guessing the question is asking for A, so that x^2 can just touch the graph of ln(x)

sparrow2 · Answer

i think derivatis must be equal : so 1/x=2ax a=1/2x^2

whpalmer4 · Answer

Well, numerically, the value of A is bounded by $$0.183935 < A < 0.18394$$

whpalmer4 · Answer

narrowed it further:
$$0.18393972058572 < A < 0.18393972058573$$

Should be a simple matter to determine the symbolic version, right? :-)

parthkohli · Answer

One thing is definitely true: when they meet, this particular curve will rise above faster than the other one.|dw:1453661430735:dw|Let's say they meet at some value $x=t$ where $t  > 0$. Their tangents must coincide there, and so their slopes must coincide.$$\Rightarrow \frac{1}t = 2At$$$$\Rightarrow 2At^2 = 1 \Rightarrow A = \frac{1}{2t^2}$$Also $\ln t = At^2$ therefore $\ln t = \frac{1}2\Rightarrow t = e^{1/2}$. Now our curve $y = Ax^2$ passes through $(e^{1/2},1/2)$  and so $A = \frac{1}{2e} $.