You Didn't Learn This In School
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- Опубликовано: 4 окт 2024
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0:51 4^(260 - x) ≠ e^ln(4)^(260-x) = e^(ln(4)^(260-x))
What you meant is
4^(260 - x) = (e^ln(4))^(260-x)
Order matters here, as powers are applied not from the bottom up but from the top down.
The log property is not log(x)^a = a log(x).
It's log(x^a) = a log(x).
You need to be careful how you write this. It's very misleading in its current format in your video.
Oloo
Every time I think about the Lambert W function, it just blows my mind. Like, it's a function with no formula, you'd think it'd be almost entirely useless. And yet it's so useful.
It can be calculated using infinite sums just like any function.
It can be represented by at least two different definite integrals over most of its domain
I mean, it is entirely useless here. What he did with the rearranging of the inside expression required solving the original equation. It’s pointless
@@dreingames9137no, he already solved the equation when he got the expression for x, he just rearranged it a bit in order to get a nice number, but the solution was already correct and finished
It's a pretty famous results that "most" functions don't have an explicit formula. That is to say, they are not expressible with a finite number of symbols and operations.
Finally, a problem involving the Lambert-W function that actually simplifies to an answer that doesn't include that function in the end. 🙂
that's like saying finally a problem involving the trigonometric functions that simplifies to an answer without pi. a solution to an equation that contains a product log is already simplified, just like a solution to an equation which includes pi, e, etc
@@reedoken6143 No, that's a terrible analogy. There are many, many videos with trigonometry problems where relatively long formulas simplify down to something elegant and even have answers with no trigonometric functions at all at the end. But virtually no video involving the Lambert-W function that I've ever watched had a problem where the final answer was something very simple like "4", they almost always have a final answer involving some algebraic function using the Lambert-W function.
Not that there's anything at all wrong with the Lambert-W function or problems that take advantage of it. But it's nice, for once, to see one that actually simplifies down to something really, really elegant.
@@Bodyknock the product log is simple and elegant though. it provides us with an exact answer to many equations for which we would otherwise need to write an approximation, just as pi, e, and all the integers do.
@@reedoken6143 You're trying to argue as if I'm saying the Lambert W function isn't useful or is confusing which isn't what I said at all. All I said was it's nice to actually see a problem that works down to something extremely simple, in this case 4. There's something satisfying about occasional problems that use a more sophisticated technique to work down to just an integer, for example, but virtually no problems like that seem to come up involving Lambert W. Which is why is was fun to have this video do just that.
So please, stop trying to talk to me like I don't understand what the function is or how it's useful or act like I'm denigrating it somehow.
@@Bodyknock I'm not trying to argue, no need to get so defensive brother. good luck on your maths journey ;)
It is interesting that the Lambert W Function plays here a similar role here, as e.g. the imaginary numbers for cubic equations, where they can cancel itself out.
That is if it has some integer solution.
Hit and trial is best , 4
lucky you
@@daniel_77.It's kinda obvious
@@mhm6421 I'm talking about that perhaps other equations may no be that obvious
You can find that solution using trial and error. However, trial and error alone cannot prove that the solution is unique. Without an actual proof, it doesn't matter how many numbers you try; you will always be left wondering if there might be some other number that works that you just haven't tried yet.
@@omp199derivative is positive for both 4^x and x
done
We can split 260 in 256 + 4
So we get,
4^x + x = 256 + 4
4^x + x = 4^4 + 4
After comparing LHS & RHS we get
X = 4
You also need to proof that there are no other solutions. Best way is to show that f(x)=4^x + x is monotonic function (sum of monotonic is monotonic) hence the equation f(x) = c, where c is real number has only one solution.
@@ВладимирНикишин-д1жbest yeah. I like to see people who TRULY understand maths like you.
@@ВладимирНикишин-д1жI mean, the video doesn’t show uniqueness either, as far as i can tell
@@teeweezevenI think it would be by the uniquness of solutions of the W function. So each step taken is an if and only if
@@shayboual1892 it's been a little while since I watched the video, but didn't they use a logarithm? That does weird things with complex numbers.
We are really not allowed to use Lambert W Functions in school exams, which is pretty lame.
Well, it's all because we didn't learn it in schools, but in your videos.
I once used it in my AP Calculus class. It appeared as a solution to a differential equation we needed to solve on an exam. We were expected to write an implicit solution, but I realized it could be solved explicitly in terms of W. I got full points
@@karolakkolo123 It is allowed only when it's the answer to a question. However, if your teacher never uses a question with Lambert W function, it's probably not allowed.
@@symmetricfivefold it just depends on the teacher
@@karolakkolo123 Yep, it totally is.
For this precise equation it's much much easier to express it as 4^x = 260-x which leads us to the intersection of two functions, and one is strictly increasing (y=4^x) and the other one is strictly decreasing (y=260-x), therefore they have only one intersection point, which is easy to find without any calculations (x=4).
Will it be accurate, won't it be recurring
@@jaisriram07007 Absolutely, you can easily proof it
@@jaisriram07007 100% accurate, because one function is **strictly** increasing, and the other one is **strictly** decreasing. These conditions are enough to say that the functions have only one intersection point
@@aram_saroyan ok
@@aram_saroyanThey also can have no intersection point.
Actually, it has infinitely many solutions when you wrote W(256ln4e^256ln4) = 256ln4 that is the principal solution but it has infinite complex solutions written as W1(x) or W2(x) or W-1(x) (the numbers are in subscript and x refers to quantity inside the function it has 2 real solutions for x between 0 and -1/e.
I just converted 260 into the form 4 to the power x + y. 260 = 256 + 4 = 4⁴ + 4.
So 4^x + x = 4⁴ + 4. On comparing, x = 4
Okay in all honesty, you take a very algebraic approach to the problem which leads to a clever use of the invert function of xe^x . But as fellow mathematician I have to expose a simpler solution (which doesn’t use a weird function that doesn’t even have a representation with the usual function).
You presented this solution, which is direct, I think in a sense of simplifying the problem to make it easier to understand the actual proof. But that lead to the use of this weird W function, and you even actually use the exponential form of a^b (e^b*ln(a)) but with that in mind using a analysis approach is way faster and easier. Take a function f(x)=4^x+x-260, with x being any real number. You can write it f(x)=e^x*ln(4) + x - 260 . The goal here id to find x such as f(x)=0 You derivate this function to obtain f’(x)= ln(4)*e^x*ln(4) +1 f’ is always positive (cuz e is define from R to R+) therefore f is always increasing. Or f(0)=-2590 (can’t be bothered to calculate it, it’s obvious). You can therefore affirm that by continuity of the function, (there is theorem name that I don’t know in English) that f(x)=0 have a unique solution. Trivially you can remark that f(4)=0. Therefore the solution to the problem is x=4 and the solution is unique.
This can be done in three lines within a rigorous form. Despite the beauty of the presented solution, this one is much more easy and elegant in its own way.
for your interest, the theorem is called “intermediate value theorem”, or IVT in short.
@@maguschuthéorème des valeurs intermédiaire TVI
That takes care of all real solutions, yes. What about the complex solutions?
the goal was to find the solution, not just show that it exists.
Soluția este unul din factorii lui 260 (2, 4, 5, 13).
* 4^x+x=260,
4^x=260-x,
4^x
To people suggesting that the host could at the very beginning just write (260) as (4^4+4) and be done with it:
That's essentially what he did except that he took an unnecessary route through the Lambert function. The host could not find a closed form solution if he had not replaced (4^260) with (4^256*4^4) (time 2:16) and then replace (4^4) with (256) (time 2:31). That's essentially the same as (actually reverse of) doing in the very beginning replacing (260) with (256+4) and then (256) with (4^4) and arriving at the solution. The use of Lambert function was wholly unnecessary since at the end he resorted to s smart breakdown of 260 into its components and he could have been "smart" at the beginning.
His approach could have some use in the general case when there is no obvious solution, though mostly obvious to almost all, which is transforming the problem of finding x in the original equation into computing the Lambert function of a constant which is tractable.
edit 3/29/2024
Here's a similar problem solved using the Lambert function:
solve for X when: X^X=4
You might think I can say X^X=2^2 which means X=2. But I can also do the following:
X.lnX = ln4
if I set X = e^Y then:
X.lnX = Y.e^Y = ln4 which means using lambert function W():
Y = lnX = W(ln4)
but if I set ln4=2.ln2 and 2 =e^ln2 then:
lnX = W(ln2.e^ln2) = ln2
and viola: X = 2
But what I was tickled to realize making up this example is that if you assume Lambert function has a closed form (which kinda does since it has a summation or continued fraction representation) then:
if X^X=A then
lnX = W(lnA)
or equivalently:
X = e^W(lnA)
I generalized it a bit. Seems like the added x term needs to be the same as the exponent.
Simple version:
A^x + x = B,
x = -W(A^B lnA)/lnA + B
More generalized:
AB^(Cx^n) + Cx^n = D
x^n =
[-W(AB^D lnB) /lnB + D]/C
Expected this to be another standard usage of the Lambert W function. Got something a bit more creative than that. Earned a like 👍
Whenever I develop pages of math sort-of like this, I often ended up back where I started.
For those of us who have worked in binary, octal, and hexadecimal for decades, the solution is immediately obvious:
4^x + x = 260
2^2x + x = 2^8 + 2^2
match the terms and x = 4
If you're going to drag out the Lambert W function just to effectively match terms, don't make the example so easy to do by inspection.
There are infinite solutions, inspection is also not rigourous
At 0:13 what does trying to build up to a repetition mean?
You have successfully convinced me that the W function is worth using, by showing a context where it can be cancelled out to arrive at solutions in preferred forms. Thank you for educating me.
I agree that the function is worth using, but I would like to point out that it was only able to cancel out because they effectively solve the original equation to do so. The original equation was 4^x + x = 260, and the step which allowed them to cancel was 4^260 -> 256*4^256. This was equivalent to finding that the original solution is 4, since they had to know that to notice how to split up 4^260 to reapply W.
i did that in a minute ig,
take mod 4 , you will get x=0 (mod 4)
then
4^(4m)+4m=260
=>256^m+4m=260, and siince 'm' is a +ve integer ,it can take only a value of '1'
so,
x=4m
=4.1
=4 (done)
At the start, I knew it was 4. Basically you ask what power of 4 is close to 260. In this case, it is 256 or 4^(4). Then you add 4 to get the 260 i.e. the RHS. So four is an ideal solution. I am in 7th grade and I solved it without using ln, e, or the lambert function.
Well that's easy lil bro
Actually, there is a noticeable thing, that there's a monotonically increasing function on the left side (because f(x) = 4 to the power of x and g(x) = x are both monotonically increasing functions), so it have only 1 intersection with the graph of the constant 260. Thus we proved that there's no other solutions besides obvious 4
clearly 4^x+x is strictly increasing, so first i would check for integer solns. clearly x=4 is a soln, which gives the only real soln
It's easy to use the guessing method.
By looking at the expression on the left side we can say that RHS must be 4 raised to some power. So by taking the sqrt of 260, the nearest small value (natural number) we can get is 4^4 = 256. Now 260 - 256 = 4 which satisfies the equation easily with x value as 4.
It only took me 5 sec to solve this.
You need to have that perfect square in mind
He is making simple equation complex 💀😳
He did, but i think the point is that this would work for every number(as long as a real solution exists at all). It's just that this one had a simple answer so he managed to cancel out the lambert function at the end
Usually a problem like this can't be solved by checking a few numbers. Naturally this one can be, but the same technique could be applied to a problem that isn't intuitive.
Simple equation ? Yes, but answering it isn't simple because it's a non standard equation
Thing is he didnt, only loked at real solutions, not complex
Nice task, but your writing of the double exponents is incorrect : (a^b)^c = a^(b c)
Your expression looked more like a^(b^c)
took me so long to figure out he was doing this and I was so confused
I have a solution that a fifth grade can solve:
so as you know that the exponent of four only has two possible last digit 4 and 6.
the last digit must be even in order to get an even number.
if it's even the last digit must be 6 and you can only add 4 or 14 or on in order to get zero in the last digit.
and as 4^14 or more is more than 260, that means it must be 4.
That would be a valid solution only if the problem specifically said "solve for integer solutions". But in this case, the solution could as well be a real or even complex number. We don't really know that until we prove it step by step
@@karolakkolo123 yes indeed, but if it wasn't an integer, the result won't be an integer probably. I'm not so expert in this but I guess you're right.
@@karolakkolo123 nope, we can already tell just looking at the equation that x=4 is the only solution, no need to make it complicated
3:05 that is a massive cancellation, well done ⭐⭐⭐ "things I didn't learn in school" is one of my favorite topics
Yeah for real
Or you can just multiply 4 on it self so many times that you can accommodate the number in 260. You will get 256, then subtraction and TADA, you will get the result, x=4.
I have another solution:
With x < 0, it's easy to see that LHS isn't an integer, so there are no solutions with x < 0.
With x >= 0, the LHS is increasing and RHS is a constant, so there is at most 1 solution, which is x = 4.
In the general case of a^x + x = b though, I think this is a nice way of solving the equation.
I have no idea what you did but I just looked at powers of 4 (which are also powers of 2) so 256 is the nearest one so it's 4
?? That's not an entirely solid method. Using the Lambert W function is though. Coming from a mathematician, that is not a surefire way (much less a good way) to solve equations like these. It's quite naive and inelegant.
What an amazing answer!
*Start by assuming that x is an integer* (this is very commonly true in RUclips videos, but may be false at times)
If 4^x + x is large, then it'll be close to 4^x (exponentials grow very fast) and therefore close to an even power of 2 which is 256 or 2^8. So, let's start there.
4^x = 2^(2x = 8) = 256 = 260 - 4 = 260 - x. Therefore, x = 4 worked and no further optimization needed.
Or one of the complex branches of W
Woking in IT (dealing regularly with powers of 2).
Took me 4 seconds to find X=4.
You should have tried with the equivalent for X=7 (X=3 being too obvious).
Finally i can convince people that the lambert function is useful
show the graphs of the equations 1:25
The function 4^x+x Is strictly increasing since it is the sum of strictly increasing functions. Therefore if there exists a solution it has to be unique. It Is easy to see that x=4 is a solution, therefore that's THE solution! Quick and rigorous.
I knew we would use the lambert w here but I did not know all those steps
I feel like im missing the point - we can find a solution immediately using the same logic as the Lambert W function from the very start, substituting x=4. Furthermore, the function 4^x+x is increasing, and so we are done. The Lambert W function is only a function because x e^x is also increasing and injective, and therefore has an inverse - i don't really see the point. Can someone help?
x=4. This is actually an easy problem. 4^x+x is monotonically increasing, from -infty to +infty as x goes from -infty to +infty, so there is exactly one real root. Let's try for integer solutions, just in case that works. Easy to see x=4 works: 4^4= 256.
You're right i didnt learn any of this shit
Math twirling
I figured out the answer immediately after I looked at it and said, that looks like it's a FOUR
Looks like 4 to me, let's see what crazy investigations will be done...
there’s a fast solution
we know both x and 4**x are monotonically increasing and continuous
so x + 4**x is also monotonically increasing and continuous
therefore it would hit y=260 at maximum one point
so if there’s a solution, that would be the only solution
it’s easy to observe x=4 satisfies the equation therefore x=4 is the only solution.
(EXTRA NOTES):
i know we would get in trouble if the right side of the equation was 259 (for example)
the only claim was that there is exactly one solution and it’s a bit less than 4
but of course the method mentioned in the video would give us the exact answer even if it is not in a simple form
we can even generalize the equation and find a generalized formula following the steps in video)
This is actually something i learned in school, the moment i saw this i grabbed my calc and did it in 20 sec
Bro tbh I finnally understood the method of Lambert thank you very much
I like this video but the solution is not a systematic way to solve such a problem because of one leap:
Eventually we get ln4 * (260-x) = W(ln4 * 4^260), and then simplify (ln4 * 4^260) to (256 * ln4 * e^(256 * ln4)), so W(ln4 * 4^260) simplifies to (ln4 * 256). But for this simplification to be systematic we'll say:
ln4 * 4^260 = ln4 * (260-c) * e^(ln4 * (260-c)), when simplified this becomes 4^c + c = 260.
Which is exactly the solution we're already trying to find. So in order to find the solution this way, we need to already know it. Essentially what Bri did, is notice this simplification in a non systematic manner, but again, you could have noticed this in the same way with the original equation.
Great video as always, but I found the pacing a bit too fast on this one.
Noted!
it's easy, (4^x)+x=260. I solved this equation and i'll show you how did i do it. 1st, 4^x is a exponental function, x is a linear function. 2nd I calculated functions values. 4^1=4; 4^2=16; 4^3=64; 4^4=256 it is a 4^x values and 256+4=260. x=4.
bro it was just a nice number what if it was 259 instead. You'd never solve it
Find a power of 4 that's a bit less than 260.
Um, 256 = 4^4
256 + 4 = 260
There's a second parantheses at 2:53
the problem with the example you chose is anyone who knows the first few powers of 2 will immediately know x is 4
I got x=4 just by plugging it in.
It isn’t really that hard, you don’t need to use complex formulas and function.
4^x + x = 260
We can rewrite 260 as 256 + 4 because the LHS has 4^x.
256 is 4^4
We get the equation as 4^x + x = 4 + 4^4
Thus, x is 4
Yes, we did not learn the W function in 1980s high school calculus
average math youtuber: Have you heard of this? It's this really really new epic thing called the Lambert W function, also have you heard of tetration, I just made a video about it?????
Me just using log in the first equation:
Me in 5 seconds after seeing the title:
4^x grows strictly monotocally, x does too, 260 is contant, so there is at most 1 solution.
4 is a solution, exercise done
So, what made you do 260 = 256 + 4 at the end? If you already knew 260 = 4^4 + 4, what's the point of trying to solve it, you already knew x=4 works 😅
I am stumped on a problem and would like help. The formula appears simple, x*sinx = 1-cosx, solve for x. I know the answer using excel, but I desire a solution. I have been trying to solve this for over a year, and have made little progress. I attempted to use the Lambert W, but the I just dug new holes without making progress. Any help would be appreciated, and I think would be a fun solution.
I suggest to look at the Desmos Graphing Calculator.
When entering both functions, you can see that they have an infinite amount of solutions, because they intersect an infinite amount of times. The only integer that happens to be a solution for x is 0.
Other solutions include 2πn, where n is an integer: So -4π, -2π, 0, 2π, 4π, etc
Funny how at 2:17 you notice 260=256+4 and that 4^4=256. This would give a much shorter proof
A single glance is all it takes to know that x=4 😂😂
or if you are a lazy person, just plot the functions and find where they intersect on the x-axis
I'd solve it old-fashioned way if we solve for integers.
1) x=0 is definitely not solution. So, x > 1.
2) x>0. take mod 4.
x = 260 mod 4 = 0 mod 4
so x =4k
3) 4^4k + 4k = 260
256^k + 4k = 260
if k > 1 we overhit, k x=4.
By trial x=4
I learnt this in college. I don't think this level of advanced maths is for school honestly
i swear to god 60% of the math vids i see on youtube involve the goddamn w lambert function
Great! I'm in 11th grade and this is really interesting!
My answer is 4. When I saw 260 and 4, I recall 256 which is 16².
The function f(x)=4^x+x is strictly increasing over R, so the equation f(x)=260 has at most one solution. Since x=4 abides the equation, it must also be all the solutions
Just by lookin,
4^x=2^(2x)
2^8=256and
4^4=256
4^4+4=260.
Now so that when the real solution is ln(3)/root(3)
I wish brilliant was free
As an experienced low-level computer programmer, I just looked at the thumbnail and thought "4". 🙂
Definitively.
2:15 the entire point of using the function is for cases that you DON'T know 4⁴+4=260 because then you literally just solved the original equation
I, an intellectual:
*immediately start counting*
Very nice 👍
0:52 why is the argument of the natural log just 4 and not 4^(260-x)?
it is the same. if you use log rules, ln(4^(260-x)) = (260-x) * (ln4)
using indices rules, (260-x) * (ln4) = (ln4)^(260-x)
@@_PK777_ the first equation you wrote is correct, but the second one is wrong. You can check this with a simpler example:
ln(2^3) does not equal (ln2)^3
@gabedarrett1301 yes but when it is e raised to the power of it, e^ab = (e^a)^b
@@_PK777_ I don't see why that matters. (ln4)^(260-x) is still very different from ln(4^(260-x))
@@gabedarrett1301he replaced 4 with e^ln(4). You can also replace 4^(260-x) with e^ln(4^(260-x))=e^((260-x)*ln(4))=(e^ln(4))^(260-x), which is the expression from the video
4^x монотонно возрастает, x - тоже монотонно возрастает => 4^x + x монотонно возрастает => не больше одного корня. Методом тыка находим 4
I just pluged 1,2,3,4 for x and concluded that x is 4 Took literaly 10 seconds
As long the equation have number less than 10000 I can solve it by just put value theorem
Let X = 4
4^4 = 256
And 256 + 4 = 260
Here's my thoughts
64....128....256 is closest to 260 so 4 difference
Let's see if 4^4 is 256 ..oh yes it is so x is 4.....moving on.
Yes that's the perfect way but you can already tell that the answer is 4 lol
x = floor(fourth_root(260))
Interesting, could you explain how you got to this result
I could just guess though
nice one
4^x + x = 260
x(4^x /x +1) = 260
x((4^x/x +1) = 13 X 2² X 5
We know x is multiple of 4 (to be an integer value), and we have two factors hence,
x(4^x/x +1) = (13 X 5) x 4
Comparing we observe,
4^x/x = 65 and x= 4 both satisfy for x=4
If someone has to do it without hit and trial this might be the best possiblility, thought by me
Again, hit and trail is the best !!
Here's a constructive criticism. IF you really want people to watch your videos and teach them something worthy of interest, DO NOT use round numbers, familiar integers, etc. I solved this one problem within 10 seconds just by looking at the numbers, and therefore, didn't need to watch what the video was about.
I did it anyway out of boredom, and actually learned something. People should watch out of curiosity, not boredom. Create curiosity with the very first image, without giving away the obvious answer, and your already impressive number of views will multiply.
Hint: 4^4 = 256.
Using my brain I'm pretty sure x=4
I solved the thumbnail in my head in 2-3 seconds LOL
So did everyone else
What if you accidentally solve it in your head within like 4 seconds of seeing the title…
15 seconds into the video and already knew x was 4
@BriTheMathGuy being "clever" (2:25) is probably the same amount(if not less) of "cleverness necessary to see x = 4 is a solution at the start :/
Ye I have no clue what's happening past the 10 second mark, good video tho
Lost me at 8 seconds in when he set the equation to 1.
bruh just trial and error. x=4