When two lines meet!! 2
Find the number of roots of equation f ( x ) = ∣ lo g 1 0 lo g 1 0 ( n x ) ∣ − e 2 n x − 1 2 n .
n is a positive real number.
Inspiration Aniket Sanghi
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1 solution
The number of roots seems to depend on the value of n. For example, I can find only one root when n=2 (at about x=10.2). Where is the other root?
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Sir, actually 2nd root will occur as a very large value as loglog(nx) is a very slow increasing graph. It will intersect the 2nd graph at a large x.
As you can see from graph, it will always have 2 roots. You can check it yourself also sir by plotting the graph. As loglognx is increasing whereas positive part of 2nd graph is decreasing, they have to intersect again.
Your graph isn't legible on my computer, so I cannot see your reasoning. WolframAlpha can find only one root. While this graph only goes up to x=100, it's clear that loglog is increasing, that the exponential is decreasing and the difference is positive. It cannot go negative no matter how large x gets. I must be missing something, but I don't know what it is.
(Apparently, Brilliant won't me put a link in a reply. Sorry)
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Links are allowed in a reply, you just need to use the correct markdown code of
[name](URL)
. I've edited your comment to reflect that (and see that you've figured it out too).
Here's a link to the photo , which you can then download and enlarge.
Can you tell me what the second root is?
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Sir, its very large, I guess above 10000000. Wolfram Alpha maybe can't find it. I can't tell you. But sir, graph is legit. You can try drawing graph. You can see sir, 1st intersection will occur between 1/n and 10/n. After that, loglog nx will go below 2nd graph. Then, as logoog nx will increase and 2nd graph will decrease(not below that line as it's asymptode), so at one point loglognx will again cross 2nd graph and go above it as it's increasing, and at a point its value has to be above 2nd graph. So, there will be 2 intersections and hence 2 roots. Now, you can prove that either there will be two or no roots as one intersection gets loglognx below 2nd graph so it has to go above 2nd graph and ther will be 2nd intersection. I hope I explained well.
Sir, I am getting 2 roots in wolfram alpha. x = 10.2 and x = 0.5 . Also sir, it takes log as log in base e. But, log is taken in base 10 here as i learnt that log is base 10 and ln is base e. I guess I'll clarify it. But, it won't make difference. Answer will still be 2.
please post a link to your wolfram page
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Sir, but wolfram alpha is very bad. Firstly, it makes graph og loglog x in -ve x side. Secondly, it is unable to detect big roots. So sir, I'll request you to please try to solve it yourself. You'll get same answer.
Fair enough. I could be wrong. However your arguments have been unconvincing. All I ask is: tell me the second root (other than x=10.2) for n=2 and I'll go away happy. You have claimed it is .5 but that's impossible. log(log(x)) is undefined at x=.5. You have also said is a very large number which was too large for Wolfram to find. Like I say, I am unconvinced.
Here's my reasoning. Your function is monotonically increasing for values above 10. The first derivative is always > 0. This happens because the exponential function is essentially constant (=1) for large values while the loglog function creeps inexorably to infinity. The values above 10.2 are all positive. At x=10^100 f(x) is 4.44 and growing. It will never see 0 again. A large root cannot exist.
If there is a small root, I sure can't find it.
Calvin., is it possible to upload images in comments? Or to post links? That would have made this discussion much less painful.
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Sorry sir if I have been rude. Actually, root is not 0.5 but close to 0.5. But. It's code is showing 0.5 as it's 0.5000000.. something Sorry again.
Ok sir. Thanku. But just a doubt(it's really a doubt). Is above reasoning valid for base 10 as my q is in base 10. I mean, really no offence. I have just started learning this topic. Thanku.
Well, I admit I was getting frustrated and lost my temper. I apologise for that, Aryan.
Everything I said up till now assumed natural logarithms. When I switch to base 10, I see your large root around 6*10^9. Whew! We can agree on something. I still don't see the small root, however. Oh well, at least I can go to bed peacefully now. :)
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Thanku sir. Sir, actually the root is 0.5000000002 something. It's rounding it up and showing 0.5.
You can post images and like in comments too sir. For link, put link name is [] and link address in () just after [] bracket.
You can also see an option below while positing questions or solutions about how to do these. I hope it helps
Thanks for the posting info. I'll try it here.
I found the small root. It's where you say it is, but it is hiding so close to the vertical asymptote of the loglog function that I overlooked it. Of course, there has to be a root on both sides, since the abs() make both sides approach +∞. So there should be two small roots, yes?. Here is a picture of what I am trying to describe.
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Ok, thanku sir. So, are there one or two or three roots? I am unable to understand. I guess you want to say that there are two asymptode of loglognx. But sir, we don't have to take other asymptode as nx should be greater than one, isn't it? So there's only one small root if it's correct, isn't it sir?
I'm embarrassed to say how much time I have spent on this with so little to show. First of all, forget everything I said about a double root. Long story short: I was using python's mpmath module which overloads the abs() function for complex numbers. It returns real numbers for logarithms of negative numbers. Eek! I should have noticed this much earlier in the game, but Wolfram apparently does the same thing so I trusted it. Doh! If this were a complex analysis problem, then it might have been correct. :)
At the risk of being wrong yet again, here is my current thinking: the smallest x in your function's domain (approximately [1/n < x < ∞]) must be positive for two roots to exist. If negative, then it can only cross the x axis once (the function is very close to being monotonic) on its way to the big root, losing the small root completely. If I'm on the right path, then n must be less than about 1.16 (numerical approximation) in order to have two roots. So it seems like I've come full circle to my original position: The number of roots depends on n.
What is your thinking now? Have I sufficiently muddied the waters that you are confused too? :) This problem beat me up pretty badly (good job!), so I doubt I will spend much more time on it. As you have probably gathered, I have little formal training, I'm just an old hack who absolutely loves arithmetic.
I do wish I could see your original sketch. My phone is too small and my eyes are too old.
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Ohk sir, I really respect you. If I understood your reasoning, actually, x can't be negative as if x is -ve, n has to be -ve for loglog nx to be defined. But, I have mentioned that n Is positive in question,so x has to be positive for it to be defined. So, no crossing x-axis and hence 2 roots, isn't it sir?
Sir, actually my whole sketch is the graph I have made in the solution.
Fletcher, you're actually on to something here. Merely knowing the shape of the graph doesn't exactly tell us the number of intersections.
What we want to do is to apply the intermediate value theorem on [ f ( a ) − g ( a ) ] [ f ( b ) − g ( b ) ] < 0 to conclude that there is at least one intersection. So by applying IVT on the regions ( n 1 , n 1 0 ) and ( n 1 0 , ∞ ) , we can conclude that there are at least two intersections.
However, there could be 2 or more intersections in each region.
To show that there is only 1 intersection will require a more nuanced argument. The condition of "
f
(
x
)
starts small and grows large,
g
(
x
)
starts off larger and doesn't grow as fast" isn't quite sufficient.
A sufficient (but not necessary) condition is that
f
′
(
x
)
>
0
>
g
′
(
x
)
. Intuitively, if one function is increasing and the other is decreasing, then they can intersect at most once (how can we prove this rigorously?).
Another sufficient (but not necessary) condition is that
f
′
(
x
)
>
g
′
(
x
)
over all
x
in the domain (not just "sufficiently large
x
"). (Can we prove this rigorously, using a similar idea as the above?).
Yup, you nailed the part of Wolfram using complex logarithm. It can sometimes be hard to tell, which is why checking the signs can be helpful.
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@Calvin Lin , sir actually I'm new to this topic, so sir is my answer correct?
Sir, my solution was that g(x) will intersect f(x) = loglog nx exactly once before 10/n as f(x) is decreasing at the point till 10/n and g(x) is almost constant. After that f(x) will start increasing whereas g(x) will remain almost constant. So, at a point f(x) will cross g(x). Then, as f(x) will increase and g(x) will remain almost constant(actually decrease), there is no third intersection. So, there are 2 roots.
Thanku sir for the explanation.
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As indicated in my comment,
- "After 10/n, f(x) will increase and g(x) will decrease", so there is only 1 intersection in that region.
- The tricky part here is rigorously demonstrating that "g(x) will intersect f(x) = loglog nx exactly once before 10/n" (Sorry for the abuse of notation.) , which is what Fletcher is picking up on. Simply saying that "f(x) will start increasing whereas g(x) will remain almost constant" isn't sufficient (E.g. f ( x ) = x / 2 , g ( x ) = sin x sortof satisfies those conditions, but there are multiple points of intersections.). One way to deal with this is to look at h ( x ) = f ( x ) − g ( x ) and show that h ′ ( x ) < 0 , to explain why h ( x ) = 0 has a unique solution in this domain.
f(x) =|loglog(nx)| - e 2 n x − 1 2 n = 0
|loglog(nx)| = e 2 n x − 1 2 n
So, we have to basically find points of intersection of y = |loglog(nx)| and y = e 2 n x − 1 2 n
Here is the graph
So, we can see that they intersect at two points. So, there are 2 roots