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
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.
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
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).
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
@@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
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.
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
@@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?
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.
@@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.
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.
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.
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 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.
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.
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? :-)
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.
@@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
@@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.)
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.
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.
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 ?
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.
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.
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.
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
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!
There is no maximum. This is evident from the process that Steve follows from 2:30 to 4:00 - just replace '3' as the index of the root with a generic n and you will always come up with the conclusion that the limit of the ratio is ∞ for any n.
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.
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
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!
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).
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 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.
If I remember correctly, there's no real number to express the order of infinity of ln (x). You'd have to have x^n with n being bigger than 0, but smaller than all numbers smaller than 0 (but no such number exists in R). Is that correct, or am I mistaken?
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?
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
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
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)
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 😊
An honest question - we work hard to rearrange the equation, so that we can use the W function and get solutions. To calculate the W function value we will need to use some algorithm like Newton's method. BUT - seriously, we could have used Newton's method already on the original problem - cbrt(x)-ln(x) = 0. So we just adjusted it to use a named function, but did we really make it any simpler?
Couple of interesting points coming out of your question: 1. Without an analytical solution to steer you, Newton (or any other root-finding method) may converge to the 'wrong' solution. For example, for starting values below 26.87, Excel's built-in Newton-Raphson algorithm converges on the zero at 6.4... rather than the one at 93.3... , and you have no way of knowing where is the zero above that (since you "don't want" to use W in this case). For the cube root the 'correct' initial guess is small enough that it's easy to get the algorithm to work. For the 12th root... not so. 2. How do you calculate the numerical values of cuberoot(x) or ln(x) to use Newton's method to find the solution? Yep, correct - using a variety of numerical algorithms (usually series) that converge to their approximate value. Since similar algorithms exist for W as well, why not use those directly? In other words, W is not such an extraordinary function. It is introduced later in mathematical studies because it relies on the student having acquired other concepts first. Just like roots, exponentials, logarithms and trigonometric functions. Quite arbitrarily (not completely), those are considered 'elementary', but there is nothing beyond familiarity that makes those 'elementary' functions truly 'elementary'.
Lambert W function explained: th-cam.com/video/Qb7JITsbyKs/w-d-xo.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?
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.
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
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
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
The fact I'm starting to solve these by myself after watching you for so many years... Thanks, dude. You're the GOAT.
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.
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
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
It can be useful when we want to compare complexity of algorithms
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
Obrigado pela explicação da função W, não a conhecia. Ótimo vídeo.
Great explanation!
Thank you for these videos
恭喜發財! Happy Lunar New Years!
Thanks! You too!
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! : )
informative video good job
Thank you!
Where can I get the integrals for you and derivatives for you table next to him?
1:20 that wow is hilarious. Great video with plot twists!
@
blackpenredpen
Can u make a video about the other LambertW functions of different indices,
lambertW(k,z), k is integer, z is complex number?
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) ?
Loved it
My mans growing a confucian beard my dude, nice.
well done Mr
After scrolling for a very long time...
4:46 "SO SATISFYING!!"
LOL!
I love this guy's teaching.
Thanks.
Lovely!
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!!
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.
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
If this has Lambert W I am leaving 😂😂
Thanks 🙏🏾
Blackpenredpen Will you please make a vedio on reimann zeta function
Is it also possible to use Newton-Raphson method to approximate points of intersection?
Which software do you use for those graphs?
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: th-cam.com/video/fo-alw2q-BU/w-d-xo.html
How to approximate sinx with prod(x - npi)(x + npi)?
7:06 “There’s always a bigger fish”
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.
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.
love the beard man hope youre doing well x
I am. Thanks. Hope the same for u too!
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? :-)
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.
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 😂
❤️
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?
How is that a solution to the equation ? How do you compute the numerical solution from that answer ?
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.)
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)?
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.
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 ?
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.
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
What does about the function represented by w
What is the software used in 4:27 the most satisfying part.
Thank you
Geogebra
Which software program are you using for plotting graphs
Geogebra, my fav!
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!
An interesting way to explain Big-O notation (kinda)
Hey BPRP, so what's the maximum n, for which x^(1/n)infinity?
There is no maximum. This is evident from the process that Steve follows from 2:30 to 4:00 - just replace '3' as the index of the root with a generic n and you will always come up with the conclusion that the limit of the ratio is ∞ for any n.
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.
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
You are great.
Does this mean that x^(1/x) eventually overtakes lnx? Doesn't x^(1/x) tend to 1?
Facts bro. I love your videos 😆😆😆
Thanks! Appreciate that
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!
in real analysis 2 and still have no clue how to solve this without the video lmao
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.
What about complex solution(s) ?
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).
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
why doing the lim of the derivative quotient is giving you the solution?
Question: For which largest degree of root of x, ln(x) will be largest at infinity?
Your question is not making sense
Just like you did a vid about x^y vs y^x could you do x tetration y vs y tetration x
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.
If I remember correctly, there's no real number to express the order of infinity of ln (x). You'd have to have x^n with n being bigger than 0, but smaller than all numbers smaller than 0 (but no such number exists in R). Is that correct, or am I mistaken?
In a sense. But now consider ln(ln(x))... and you see why people came up with non-standard analysis.
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?
the wofram screen that you put up at the end is too fast to see . . .
I would have preferred to compare log to x^(1/n) where n is a positive integer
where are you from bro?
what software did you use to plot the graphs?
I think it's geogebra.
Yup
Great job... it’s cool... 😎
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
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?
How about x^(1/x) ?
i didn't know about the w-function. got stuck at e^3=(e^x)/x :(
Does this mean that ³√(∞) will give a larger ∞ than ln(∞)?
What is the maximum of ln(x)-x^1\12
Can I get a heart from Blackpenredpen???
Huge appreciation to you from India 🇮🇳🇮🇳
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
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
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)
Hello, I have interesting question for you: how to solve lnx =sinx?
I don't think there is an analytical solution to it.
It can be solve doing ln(x)-x^1/3=0. U have to look at it like a function
didnt even sa thank u for derivatives at the side but he is saying welcome🤣
In a way isn't ln(x) the function that is asymptotically smaller than all of x^(1/n)?
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 😊
Where were you during my uni days?
An honest question - we work hard to rearrange the equation, so that we can use the W function and get solutions. To calculate the W function value we will need to use some algorithm like Newton's method. BUT - seriously, we could have used Newton's method already on the original problem - cbrt(x)-ln(x) = 0. So we just adjusted it to use a named function, but did we really make it any simpler?
Couple of interesting points coming out of your question:
1. Without an analytical solution to steer you, Newton (or any other root-finding method) may converge to the 'wrong' solution. For example, for starting values below 26.87, Excel's built-in Newton-Raphson algorithm converges on the zero at 6.4... rather than the one at 93.3... , and you have no way of knowing where is the zero above that (since you "don't want" to use W in this case). For the cube root the 'correct' initial guess is small enough that it's easy to get the algorithm to work. For the 12th root... not so.
2. How do you calculate the numerical values of cuberoot(x) or ln(x) to use Newton's method to find the solution? Yep, correct - using a variety of numerical algorithms (usually series) that converge to their approximate value. Since similar algorithms exist for W as well, why not use those directly?
In other words, W is not such an extraordinary function. It is introduced later in mathematical studies because it relies on the student having acquired other concepts first. Just like roots, exponentials, logarithms and trigonometric functions. Quite arbitrarily (not completely), those are considered 'elementary', but there is nothing beyond familiarity that makes those 'elementary' functions truly 'elementary'.