comparing cbrt(x) vs ln(x)
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- Опубликовано: 25 июл 2024
- Which function is eventually bigger: the cube root function or the natural logarithm function? Visit brilliant.org/blackpenredpen/ to start learning and exploring fun Brilliant courses. That link also gives you a 20% off discount on their premium subscription.
In this video, we will compare radical function and logarithmic function: which one is eventually larger? We can take the limit as x goes to the infinity of their ratio. With L'Hospital's Rule, we can see that the cube root of x should be eventually larger. So can we find the other point of intersection?
Here's a lecture on the Lambert W function: • Lambert W Function (do...
0:00 sqrt(x) vs. ln(x)
0:39 cbrt(x) vs. ln(x)
4:27 the most satisfying part
4:48 learn more interesting math on Brillaint!
5:47 solving cbrt(x)=ln(x)
9:45 different branches of W(x)
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Thank you,
blackpenredpen
Lambert W function explained: ruclips.net/video/Qb7JITsbyKs/видео.html
Which would dominate if it was x^1/n where n goes to infinity?
Can you please solve this question
I'm not asking for a video
Just solution
If its possible, anybody please help i can't find it on internet
2014^x + 5^x = 2008^x + 11^x solve for real values of x
By hit and trial i know 0 and 1 are the solution but i want to know if there's a method to properly solve it
Where are you from?
In reality you haven't solved the equation but you have rearranged it and called the solution "Lambert function". It is called circular reasoning. In reality your solution is no different to the solution x=(lnx)exp3 put in Walpha nad find the solution. This problems should be called rearange equations with the help of the Lambert function.
@@igorkarlic2297 you can calculate W(n) by hand if you really wanted to
Root of a tree vs log of a tree
Wood is wood, a valuable material, in particular with the beavers.
Root will win eventually
Got the joke.
@@Apollorion Yes, that's why you use the Lambert Wood(x) function.
Which is bigger, root(tree(3)) or log(tree(3)), where the index of the root and the base of the log are both G64?
Bro ..i am following you since 3-4 years.. whenever i see your old videos., It makes me emotional as it reminds me of my childhood... things have changed in these years...most specifically what i remember is
BLACK PEN RED PEN YAY
Awww thank you
In this case, L'Hopital is just a fancy name for "the function which eventually increases faster will eventually win".
Very valuable comment!
Not all limits will go to zero
Right. Just take the derivative until you get a linear term or a constant. Whichever is larger will always win.
I actually thought about using the inverse functions of nthroot(x) compared to ln(x) and seeing that e^x is going to be bigger than any x^n eventually and this leading to the inverse being smaller than any nthroot(x).
Same
Wow thanks never thought of it in this way. What I did was take the derivative of both and realised x^any postive interger always has larger slope than log for huge numbers, which is kinda similar to the video. But your method is way better since we cant say anything by just the first derivative, higher derivatives matter too
Yea that's what I was thinking!
What about infinith root?... Does that tend to Ln X?
@@adammckay7335 putting aside the fact that you can't raise something to an infinity power... It is still the case that x to ANY POWER is going to be smaller than e^x over time, so even though ln(x) may look to be above for millions and millions down the number line, the root graph will eventually win
The fact I'm starting to solve these by myself after watching you for so many years... Thanks, dude. You're the GOAT.
Only for blackpenredpen would I stay up at 3 AM to watch a new video!
Appreciate that!
It's actually 4:16PM here, afternoon, in India
3:54 in Germany 😅
Now here is 03:21...between midnight and dawn..
It's now 3 AM in my country
6:26 that ".....thank you so much" reminds me of Dr. Peyam's "thanks for watching" for some reason xD
thanks for commenting! : )
In algorithm design they want your algorithm to be O(ln(n)), but your solution is O(n^(1/12)), technically it isn't O(ln(n)) but in more common small cases, it's actually better :) 4:30
You'll basically never reach that point so no worries 😃
@@pbj4184 It very much depends on the problem. The critical value for 12th root is somewhere of the order of 10^20, which is pretty small for some applications. Prime factor decomposition or primality tests, anyone?
@@dlevi67 i'd love to see an algorithm that's actually O n^(1/12) though XD but yeah it's not really that common to find values over 100 quintillion
Why is no one talking about how amazing Brilliant is? I've seen hundreds of videos which Brilliant have sponsored and I've never seen any comment about Brilliant. Underrated.
Brilliant is indeed awesome
People probably think it's a scam because they've sponsored so many videos.
x^1/e and ln(x) do a really weird thing at x=e^e they touch at that point but x^1/e is always greater otherwise
My initial thought was to create a new function that represents the difference between the two functions, find the derivative of this difference function you will notice regions where the difference is increasing or decreasing.
Nice! But do you mean integration of the difference function?
@@DarkMage2k i was talkking about just analysing the derivative function of this difference function, if the derivative is negative then you know the two functions are moving away from each other, the converse is true.
@@nathanobiekwe6836 oh I misunderstood that area thing. I thought like area under the curve
Integral nice
It can be useful when we want to compare complexity of algorithms
Obrigado pela explicação da função W, não a conhecia. Ótimo vídeo.
恭喜發財! Happy Lunar New Years!
Thanks! You too!
Great explanation!
thanks to blackpenredpen I learned about the W-function and solved this equation in less than 40 seconds
You're welcome and I am glad to hear! : )
Thank you for these videos
Thank you for showcasing the secondary branch. Maybe make another video revolving around the branches of the Lambert W function? Feels like you just kinda threw it out there, yeah just put a minus in here and you'll be good.
informative video good job
1:20 that wow is hilarious. Great video with plot twists!
This is really interesting. Its also cool that you generalised that x^(1/n) is always bigger than lnx as x goes to infinity from "The List". I just wonder if it's possible to prove it. Still very useful.
Yes this is how I introduce “the list” to my calc students.
exponentials grow faster than polynomials. hence the inverse of polynomials will grow faster than of the exponentials. try to think like that, it probably will make sense.
@@hattapalkan8395 thank you. I think i understand now.
@@Ninja20704 glad i could help!
To prove the limit of x^a/ln(x) when x goes to infinity is 0 (and with that, that x^a Gets arbitrarly larger than ln(x) ) , do the change of variable x=e^y
My thinking is, e^x is always bigger than any polynomial when x is sufficiently large, because the Taylor series of e^x is of a higher degree than any polynomial and its coefficients are all positive. Therefore, the inverse of e^x (i.e., ln(x)) will always be less than the inverse of a polynomial (such as a square or cube root of x) when x is sufficiently large.
smart
that's so overkill, I love it
There can't be a bprp video without Lambert's W(Product Log) Function. He is trying hard to make this a part of the course!!
Loved it
Thank you!
I love this guy's teaching.
Thanks.
Lovely!
@
blackpenredpen
Can u make a video about the other LambertW functions of different indices,
lambertW(k,z), k is integer, z is complex number?
well done Mr
love the beard man hope youre doing well x
I am. Thanks. Hope the same for u too!
Am I high, or can't you just say that e^cbrt(x) will eventually outpace e^(ln(X)) = x thus at some point cbrt(x) must be greater than ln(x) ?
Blackpenredpen Will you please make a vedio on reimann zeta function
It's been too long since I mathed. ;-) I forgot about "ln" = natural logarithm, so at first thought l and n were variables from the video thumbnail. Then seeing "cbrt(x)=ln(x)" in the actual title made more sense...but I still needed to listen to the explanation.
Love your video! (from Malaysia)
I just wanna seek your advice, I always love to explore different topics of maths and I wanna know where can I start if I'm interested to learn about Pure Maths, which websites/videos or books I can refer to?
I'm no bprp but this might help: ruclips.net/video/fo-alw2q-BU/видео.html
Another easy way to do it: For n in N, we want to compare x^(1/n) and ln x when x tends to positive infinity. Taking ln for both, then it is to compare 1/n * ln x and ln ln x. If we fix n, then the magnitude of 1/n is irrelevant. We may as well just say we are comparing O(ln x) and O(ln ln x). Since x > ln x for x > 0 (which can be verified by computing the minimum of f(x) = x - ln x), we have ln x > ln ln x by the fact that ln is increasing. Thus, x^(1/n) >> ln x for each n in N.
Nice
Wow! Really 😎
To be honest, the magnitude of 1/n is not so irrelevant. In general, proving that f(x) > g(x) for all x in a certain domain D doesn't really imply that, m*f(x) > g(x), neither in the same domain D nor in a domain C smaller than D (where m is a positive real number). For example 2x > x for all x > 0, but for m = 1/4 we have m*2x = x/2 < x for all x > 0. Therefore proving that ln x > ln (ln x) for all x > 1 is not sufficient to prove that 1/n * ln x > ln (ln x) for all x > 1 and for each n in N. The last inequality is indeed false, in the sense that for n = 3 we have just seen in the video that the function cbrt (x) is eventually bigger than ln (x), but not for all x > 1.
Where can I get the integrals for you and derivatives for you table next to him?
After scrolling for a very long time...
4:46 "SO SATISFYING!!"
LOL!
Exponential always beats polynomial in growth.
So the inverse function of the exponential, which essentially transforms back the exponential to x, must be smaller to cancel out the bigger exponential function for large x.
I remember being told point-blank that any positive-valued power (i.e., exponent is greater than 0) eventually outgrows the log function. Now I understand why!
I think of them as inverses of e^x and x^3. If you take the natural log of each, you’re left with x and 3lnx. Taking the derivative of each, you get 1 and 3/x. The derivative of the second continuously decreases, so it will eventually lose its lead.
I did it by substituting in u, where x=e^3u, and then the question become whether e^u or 3u is bigger, and of course an exponential function grows faster than a linear one.
This one is great! Is there a name for this method or did you come up with it yourself? Also how did you specifically find f(u) for x
But this substitution has a miatake because cbrt of x can be negative while e^3u is always +ve..
@@pravargupta6285 u doesn't necessarily have to be an integer, take complex values and take the modulus
An interesting way to explain Big-O notation (kinda)
Thanks 🙏🏾
I just picked some big number n, where n/3 is an integer (like 900). e^900. cbrt(e^900) = e^300. ln(e^900) = 900. e^300 > 900 clearly. Therefore cbrt(x) grows faster.
Sometimes its nice not to be technical
I think this is because the graph of the inverse of a function is the graph of the function but reflected along the line y=x. Therefore since exponential functions get bigger than polynomial functions normally, the reflection makes the root functions get bigger eventually and the log functions relatively smaller.
Really cool trick with the Lambert W function.
Will you do an extension video for nth root of n vs lnx?
like how?
@@blackpenredpen I think he wanted to say nth root of x. Basically if there is some n which makes the ln bigger than the root. But looking at the limit i think this is never gonna happen xd
Sorry, I meant Xth root of X, as in x^(1/x).
@@blackpenredpen Could be interesting to find the value a (0
@@danielnieto7714 In my humble opinion, I think there is no sol of n that will satisfy your statement. Like I tried to conpare the nth root function and natural log function that i got lim x -> inf ( x^1/n ÷ ln x ). After doing the L 'Hospital's Rule, I get x^(1/n) ÷ n, and i plug this in walfram alpha. It shows that no n > 0 satisfy the result of getting 1. So I think there won't be a nth root that will actually eventually be smaller than ln. (I don't know if my calculations is wrong tho.)
Great video! Could you please explain how are the branches of W ordered? 🙏🏻
Hi! Did you came from the bprp fast channel’s newest video?
I was watching your videos before 2 years
And now it suddenly came in my recommendation
And i came to know you have grown a beard 😂
❤️
This is a part of Algo. Bro I suggest you to make videos on computational complexity. Teach theory of computation
This is standard function analysis, nothing particularly connected with algorithms. Not that computational complexity isn't interesting, so a vote from me for that too!
Facts bro. I love your videos 😆😆😆
Thanks! Appreciate that
7:06 “There’s always a bigger fish”
The first question can also be anwsered by looking at the inverse of the functions (x^3 and e^x) and asking which one grows the slowest. x^3 grows the slowest, thus it's inverse 3√x grows faster than ln(x).
You are great.
If this has Lambert W I am leaving 😂😂
Is it also possible to use Newton-Raphson method to approximate points of intersection?
My mans growing a confucian beard my dude, nice.
Does it mean, if I didn't use 3rt or 2nd root of x, but some parameter, e.g. "s," so if aftel l'hopital I find such "s" that that ratio would be 1, I found an "s-th" root of x which is the smallest one from which those two functions always cross out each other? :-)
ln(x^2) is define on R - {0} then ln(x^2) = 2 ln(x) which is now define on ]0 , +infini[ Why there is a diff between this two domains defintion for the same function ?
Great job... it’s cool... 😎
How is that a solution to the equation ? How do you compute the numerical solution from that answer ?
How to approximate sinx with prod(x - npi)(x + npi)?
Just like you did a vid about x^y vs y^x could you do x tetration y vs y tetration x
What is the software used in 4:27 the most satisfying part.
Thank you
Geogebra
Does this mean that x^(1/x) eventually overtakes lnx? Doesn't x^(1/x) tend to 1?
I would have preferred to compare log to x^(1/n) where n is a positive integer
Could this be generalised as
Lim as x->infinity of (nth root of x)
is always bigger than
Lim as x->infinity of (log n of x)?
What does about the function represented by w
another way to see dominance of x^d (d>0) over lnx: substitute t=x^d, get t >> 1/d * ln(t) = ln(t^1/d) (since t>>ln(t))
Which software do you use for those graphs?
in real analysis 2 and still have no clue how to solve this without the video lmao
Let's say the question was (3rd root of x) wich is also x^1/3 and let's say u were to say x^1/3 = ln(x) how would u solve for x , I had a lil problem with this one but it was just me making a mistake and the answer I got was {x = -27 LambertW (-1/3 )^3 , x = -27 LambertW (-1, -1/3)^3 }
It would be nice if u made a vedio on this so I can also confirm my answer that would be much appreciated 😊
Which software program are you using for plotting graphs
Geogebra, my fav!
You can do it faster if you differentiate both cbrt(x) and ln(x), and you get cbrt(x) = 1/3 int x^(-2/3) dx and ln(x) = int x^(-1) dx. -2/3 > -1, thus for some x in R+ we have cbrt(x) > ln(x) (via comparison)
Exponential growth is well known to be always bigger than any power (polynomial) growth eventually - e.g. a^x >> x^n (for any constants a and n, x -> infinity). So, the reverse is also true. ln(x)
Quick question: other than brute forcing this by looking at the graphs out to an arbitrarily large x-value, is there any way to tell that these two functions only have the 2 intersections?
Differentiate y = cuberoot(x); also differentiate y =ln x.
Observe that after that second intersection cuberoot(x) will always have a larger gradient than ln x, so there can be no more turning points.
What about complex solution(s) ?
Why do you use a fish as the argument of the Lambert fn?
Is this a common notation or a BPRP invention?
It's his invention. The point is that it doesn't matter what you put in, but the argument of the product and the log has to be exactly the same.
Can I get a heart from Blackpenredpen???
Huge appreciation to you from India 🇮🇳🇮🇳
why doing the lim of the derivative quotient is giving you the solution?
Haha I feel like it's also cuz the inverse of the two would be an exponential and the other would he a polynomial... and so exponential should always be larger than polynomial right?
This can be generalized to any root. The nth root of x, in its rigorous form, is defined to be equal to exp[ln(x)/n]. So exp[ln(x)/n] = ln(x). This is equivalent to -1/n = [-ln(x)/n]·exp[-ln(x)/n]. This implies ln(x) = -n·W(-1, -1/n) or ln(x) = -n·W(0, -1/n). If you choose x such that ln(x) > -n·W(-1, -1/n), then it will be apparent that exp[ln(x)/n] > ln(x), hence proving the root grows faster.
Another form to prove it is by realizing that ln(x) = exp(ln[ln(x)]), so comparing the nth root to the logarithm is equivalent to comparing ln(x)/n to ln[ln(x)], and with this, it does become somewhat more apparent that the formed grows faster.
Dont know if my method is correct or no.
First Take two functions y=x^3 and y=e^x
Since the second functions is exponential and will take over x^3 at some point so e^x is bigger than x^3
So the inverse of e^x is smaller than inverse of x^3
Even though easy, this is pretty nice example of how simply graphing something can lead you to false conclusion. Just remembering that logarithm is the inverse of exponential and root is always polynomial no matter the size of the exponent, we will know that logarithm is always going to eventually be smaller, no matter the exponent of the root. It looks weird, if you take that to extreme and numbers will of course get huge. Similar thing happens with comparisons like ax^b vs c^dx where the exponential always wins (I think, maybe proof one way or another could be an idea for video ...) given that all the constants are positive and c>1.
In fact I wish Steve had pointed that out - if instead of "1/3" you take "1/n" at 3:57 (or e^(-n * W¹(-1/n) at 9:03) this is analytically clear.
(sorry for the superscript ¹, but YT editor is what it is)
Question: For which largest degree of root of x, ln(x) will be largest at infinity?
Your question is not making sense
You can simplify the answer at 9:00 like this to get -27W(-⅓)³:
e to the power of -3W(-⅓) = e^W(-⅓) to the power of -3
e^W(-⅓) = -⅓ divided by W(-⅓), or -1 / 3W(-⅓), so our answer is equal to -1 / 3W(-⅓) to the power of -3, or -27W(-⅓)³
i didn't know about the w-function. got stuck at e^3=(e^x)/x :(
so easy to solve. Reduce[x^(1/3) == Log[x]]
It can be solve doing ln(x)-x^1/3=0. U have to look at it like a function
What happens when we have this: f(x)=x^(1/a), a=lim (t->infinity) t and we want to solve lim (x->infinity) f(x)/lnx????? Is this even a thing?? Is it computable?? And if it is how we can solve it?? Does the solution tell us anything?
I have two question regarding this video:
1) are there always TWO solutions (intersecting points)?
2) If there are always two solutions, what is the biggest difference between ln(x) and x^(1/n) for x being inside these two intersection points
Differentiate the difference function and find the max using Fermat theorem
@@Yougottacryforthis thank you
How about x^(1/x) ?
I can almost always tell now when the Lambert function will come into the picture.
It's pretty much whenever x is in a log or exp but also outside of said function
this problem was easy. The derivative of ln x is 1/x and the derivative of cuberoot(x) is (1/3)(1/x^(2/3)). 1/x gets smaller faster as x approaches infinity, therefore the slope of lnx gets less steep more quickly, and thus cuberoot(x) is the winner. In fact, any root of x will win because the derivative will always have a power of x less than 1 in the denominator.
if instead you tried the e-th root, they will only intersect at one point, at (e, e^e)!
Wait am I missing something? if you plug in e^(1/e) to to x^(1/e) you get e^(1/e^2) and plug it in ln(x) you get 1/e.. so the value of both functions aren't equal at x=e^1/e then how is that the intersection point. Sorry if I am dumb but this seemed sus
@@anonymousfry y value will be e and the x value will be e^e, sorry
Hello, I have interesting question for you: how to solve lnx =sinx?
I don't think there is an analytical solution to it.
cbrt(x) = log(x), take z^3 = x so that z = 3log(z), rearrange to get (1/z)log(1/z) = -1/3 which means log(1/z) = W(-1/3). Solve for z and cube both sides to get x = z^3 = exp(-3W(-1/3)). The "trick" is to realize that xlogx = W^-1(logx).
what software did you use to plot the graphs?
I think it's geogebra.
Yup
ln(x) is the inverse of e^x, while nth root of x is the inverse of x^n.
We know that e^x (exponential function) will eventually always be greater than x^n (polynomial function). Hence the inverse ln(x) will eventually always be smaller than x^(1/n).
Where were you during my uni days?