In my school teacher said something like “pick one thing from math and show it” and one chose fractions with different denominators and teacher said to mark numerator and denominator and the arrow looked like integral so here it is numerator S 2/3+1/3 denominator (the bottom half of fraction I’m bad at math in english)
How to make 4:30 sec long video tutorial Hello guys welcome to my youtube channel, today we are going to go through this mathematical problem, from a famous University questionaíre which took me quite a while to figure out how to solve.. now before we jump into the problem itself, like subscribe share and tap that bell icon as it helps me a lot to get motivated about making of these videos. Now to the problem itself => There's a link to it in the description, you can click and get your desired answer. Thankyou for watching this video guys, I'll see you next time. Until then take care and bye and don't forget to comment how much you liked my video. You can give your feedback and suggestions too to improve the channel as I'm quite new to do these type of content. Peace out.
in order to evaluate the trivial negative solution -2, we have to consider that w(x*lnx)=lnx; in fact we have: 2/ln2 * w(-ln2/2=2/ln2*w(0.5ln0.5)=2/ln2*ln0.5=-2 In the case of point, this is relevant to the first branch w0 of the inverse lambert function, i.e. for w>-1; in fact we got in the previous calculation: w(0.5ln0.5)=ln0.5=-ln2>-1. On the other hand, for the second negative solution, i.e. -4, it is necessary to employ the second branch w1 of the inverse lambert function, i.e. for w
Instead I would have done the following: We can take the ln as both sides are positive: ln(x^2)=ln((1/2)^x) 2ln(x)=xln(1/2) (1/x)ln(x)=1/2ln(1/2) Multiply by negative 1 (1/x)ln(1/x)=1/2ln2 W(ln(1/x)e^(ln(1/x)))=W(1/2ln2) -ln(x)=W(1/2ln2) lnx=-W(1/2ln2) x=1/(e^(W(1/2(ln(2))))
The lower branch is the one I found first (-2)^2=2^(-(-2)). This problem is the same as finding x for x^2=2^x where 2 and 4 are quite obvious only the third solution is a bit tricky, all just with a negative sign.
@@sherueatyourbestfriend6791 2:39 Here in the first eq. you can't apply W -Function But he nicely took the square-roots on both sides and multiply by ln(2)/2 That was what I like.
I spotted the solution -2, then solved using the Lambert W0 function to get 2 solutions (one positive and the other negative which I assumed to be -2), but alas didn't spot the obvious solution -4 nor did I think of using W-1. So I really appreciated your careful approach!
I am a korean and your explanation is easier than learning in korean. So, can you make a video of defining ln and e ? It will be really helpful for me. If there is, i really want the link. Thanks.
Nice video! You seemingly ignored the other infinite answers to the equation eg. ~(-7.6545+11.9537i) (from branch 1 of +ve value in W) that come from the lambert w equation, would have been good to at least mention these as the possible complex solutions I think!
Complex solutions are only relevant in engineering. He is tackling it from a pure maths perspective, at least this is my interpretation of his solution.
@@applealvin9167 bro I am in 8th grade so I don't even know what's a function i've tried to learn it but I couldn't thats why I asked what's the basics
@@jasnoor8-d-155 functions are how you turn the shape of lines on a graph into a set of numbers, there are range of function patterns like a simple straight line or a "parabola", to start of with functions you can watch some videos on straight line functions and go from there.
@@jasnoor8-d-155 a function is simply something that changes the input. If your function f(x) = x + 2 , f(0) gives 2 for example. By the way, with the Lambert W-function you can manipulate functions to make them easier to solve/integrate etc. If you really are in 8th grade by the way, I wouldn't recommend watching videos about the Lambert W function as it is quite advanced stuff. Instead I would just focus on the basics about functions, and all kinds of things you do with them.
in equations like square root of a^x I'd use some sort of property with the absolute values even if the exponent is even or not and get rid of the square root. or other roots if I'm not wrong. I just get the fact that |x|² = x² edit: this property works for any exponent raised on absolute value so having some correct nth root can further simplify things
well X² = 2^-x .... apply ln to both sides -> 2 ln ( | x |) = -x ln ( 2 ) --> ln(|x|) / x = -ln(2) / 2 --> ln(|x|) / x = ln(| - 2 |) / (-2) thus comparatively x = -2 for the positive solution ....... Mr Lambert need your Help please
The square root of a value means what number squared equals that value. For example, sqrt(4)= +/- 2 because (2)²=4 and also (-2)²=4. There are two correct answers. There really isn't such a thing as the "positive square root" function, but if you mean the sqrt(+1), well that seems to imply you would expect a negative answer to the sqrt(-1). But that is the definition of i, the imaginary number, and has no bearing on the sqrt(+1).
@@ajreukgjdi94 positive square root function is y=sqrt(x), so if we put x=4 we get corresponding y=2 and not (+-2). Y^2=x is not a function but when we take sqrt in both sides we get 2 branches 1.)y=sqrt(x) which is positive square root and 2.)y=-sqrt(x) which is negative square root. Take any calculator or ask any learned mathematics teacher sqrt(4) is always written as 2 . However if we have a variable inside the sqrt function (say t^2) then the output is +-t.
Sorry, math noob here. Is W really the best way to express it? I mean, does the inverse function really can't be expressed and simplified using elementary symbols we know?
W is used to solve problems like this, there is indeed no other inverse (in terms of the elementary symbols I think you mean). But really, compare this to using a square root to solve an equation like x^2 = 7. You are happy to say one of the solutions is the square root of 7, even though in some way that doesn't tell you the value either. You can of course express it as a decimal, but that's the same for the W function. The square root is just more familiar, in some sense more basic, and it has been used for a longer time.
Solution of this kind of problem depends on constants in it. So there is only constant 2.1/2 is just a variation. Put -2 on both aide aand get 4on both side.
We're not saying that this is an *identity*, i.e. that it is true for all values of x. So, to pick an example of an identity sin 2x = 2 sin(x) cos(x). We can choose any value of x, and the relationship will be true. Or, if you're not familiar with trigonometric functions yet, consider (a + b)(a - b) = a² - b². Again, whatever values of a and b we pick, the identity holds true. Technically, one should write (a + b)(a - b) ≡ a² - b², and the ≡ symbol (triple bar) makes it clear that it is an identity, not an equation, but in practice the simple = (equals) sign is used. Identities are useful where we want to change the form of an expression into some other form, that makes it easier to manipulate or solve. We can change the form because we're comfortable in the knowledge that changing the form doesn't change the substance of the expression or change the values of x that make the equation true. An *equation* on the other hand is usually only true for some specific value(s) of x, or whatever variable it is expressed to be in. So, if we have, say, an equation 2x + 5 = 23, we can rearrange and solve and get a solution, x = 9. The equation doesn't hold for any other value of x. As equations become more complicated, of higher order (quadratic, cubic etc), or involve trig functions or countless other types of function, equations will have multiple solutions. As a very simple example, x² = 9 has two solutions, which are x = 3 and x = -3. And so on. Indeed, some equations will have *no* solutions because there are no values which make them valid on both sides. That equation may have been artificially created as a problem question, just to test a student's ability to solve, or it may come about as part of a calculation or proof based on certain assumptions, which are then shown to be impossible. Sometimes there will be no *real* solutions, but delve into the world of complex numbers and we can find solutions that work. These may or may not have some tangible significance in the real world but can exist as mathematical ideas nonetheless. Hope that helps.
I would undo both of the initial exponents to get x^(1/x) = +- .5^.5. Take the reciprocal of both sides and we get (1/x)^(1/x) = +- sqrt(2), but from here idk
It is defined simply as the inverse of the function x → xe^x. If that’s not satisfactory, then think about this: How would you define the square root? Well it’s just the inverse of the function x → x². It’s the same idea.
It isn’t necessary for the simple solutions (-2,-4), but it’s hard to find a method for finding them that isn’t akin to guess and check. For the third solution, yes, it cannot be expressed in terms of elementary functions, so the lambert W function or some analogue is necessary:
i paused the video and tried to solve this equation. i took log on both the sides and then i cross multiplied the result and i got lnx/x=-ln2/2 so x=-2
0^0 is considered to be *indeterminate*. Precisely what that word means (and how, for example, it is different to "undefined") is probably, with respect, beyond the level of this question. It's to do with calculus and limits and by all means do some further research into that, but we can look at this in a rather simpler way. We have two "rules" that we are trying to apply at once, and each rule is giving us a different answer. One the one hand, we have a rule that says zero raised to any power, is zero. That's not hard to envisage. We can start with zero, multiply it by zero over and over again, and still get zero. [Side note - when we increase the power we're "multiplying up", when we decrease the power we're "dividing down", so you might see that going from 0^1, which is 0 (anything raised to the power 1 is itself) to 0^0, means dividing by 0, and if you are familiar with that idea you will know it causes its own problems] On the other hand, we have a rule that says anything raised to the power 0, equals 1. Now that is possibly *less* obvious and you may or may not yet be satisfied that it is the case, but it is a universally accepted rule. Again, you can do a little research on *why* that is the case, and that's a slightly different question. So when we look at 0^0, we have one rule that says the solution should be 0, and another that says it should be 1. We can't stick to one rule without breaking the other, so we call it indeterminate. It's like being on a road where one sign says "Speed limit 20mph" and a sign next to it says "minimum speed 30mph". We can't have both rules at the same time without breaking one of them.
@@tanelkagan 0 isn't a negative integer, so it wouldn't be dividing. ^0 would read that the base is multiplied by itself a total of 0 times. And so there should never be a rule suggesting that ^0 should ever equal 1. Further, 0^0 cannot result in a number greater then 0 when multiplied by a multiple of 0. So even if there were a rule suggesting that x^0 could equal 1, it would not if x=0, because of basic multiplication principles. I've looked into the reasons why x^0 could equal 1, and they all make less sense then a negative integer having a square root. It's a mathmatical impossibility. The question is how are exponents defined? You multiply the base a number of times indicated by the exponent value by itself. So in multiplying 0 by itself no times, I'm left with 0.
I bet there's gonna be a Lambert W function in this video. Like, it's an algebra equation, and suddenly a Lambert W function appeared out of nowhere. Kinda hate it. Edit: I was actually correct.
🎓Become a Math Master With My Intro To Proofs Course! (FREE ON RUclips)
ruclips.net/video/3czgfHULZCs/видео.html
no
H- how is this comment showing 9 days ago when this damn video was posted 1 hr ago bro
@@nice_mf_ngl I think its not glitch....its glich
@@nice_mf_ngl the video was private when it was published 9 years ago, then it become public less than hour ago
In my school teacher said something like “pick one thing from math and show it” and one chose fractions with different denominators and teacher said to mark numerator and denominator and the arrow looked like integral so here it is
numerator
S 2/3+1/3
denominator (the bottom half of fraction I’m bad at math in english)
I absolutely love that you make no intro and go straight to the problem. Wish more youtubers did that
How to make 4:30 sec long video tutorial
Hello guys welcome to my youtube channel, today we are going to go through this mathematical problem, from a famous University questionaíre which took me quite a while to figure out how to solve.. now before we jump into the problem itself, like subscribe share and tap that bell icon as it helps me a lot to get motivated about making of these videos. Now to the problem itself =>
There's a link to it in the description, you can click and get your desired answer.
Thankyou for watching this video guys, I'll see you next time. Until then take care and bye and don't forget to comment how much you liked my video. You can give your feedback and suggestions too to improve the channel as I'm quite new to do these type of content. Peace out.
@@aaaaaattttttt5596 🤣🤣
The passion of maneuvering these insane yet beautiful equations is a great example of critical & meticulous thinking of a human mind. Absolute Gem.
Well said bro
I'm glad you enjoyed it! Have a great day!
"insane" is always beautiful.
@@Thomas_Name Yes only a handful of people would understand it.
@@BriTheMathGuy as a 9th grader it goes right ahead my head
Thanks!
This is the second video in a row with a length of 4:28.
Upload date: 2/14
4:28/2 = 2:14
*COINCIDENCE? I THINK NOT!*
🤯
Impeach Wolodymyr Zelens'kyj!
in order to evaluate the trivial negative solution -2, we have to consider that w(x*lnx)=lnx; in fact we have: 2/ln2 * w(-ln2/2=2/ln2*w(0.5ln0.5)=2/ln2*ln0.5=-2
In the case of point, this is relevant to the first branch w0 of the inverse lambert function, i.e. for w>-1; in fact we got in the previous calculation: w(0.5ln0.5)=ln0.5=-ln2>-1.
On the other hand, for the second negative solution, i.e. -4, it is necessary to employ the second branch w1 of the inverse lambert function, i.e. for w
*how to summon a demon*
Instead I would have done the following:
We can take the ln as both sides are positive:
ln(x^2)=ln((1/2)^x)
2ln(x)=xln(1/2)
(1/x)ln(x)=1/2ln(1/2)
Multiply by negative 1
(1/x)ln(1/x)=1/2ln2
W(ln(1/x)e^(ln(1/x)))=W(1/2ln2)
-ln(x)=W(1/2ln2)
lnx=-W(1/2ln2)
x=1/(e^(W(1/2(ln(2))))
I think you can't use log-rules in your second step since you don't know if x is positive.
@@thatdude_93 yep
ln(x^2)=2ln|x|
@@thatdude_93 he can use it since ln(x^2)=2ln(|x|)
You forgot to put |x| so there is in reality 3 solutions
The lower branch is the one I found first (-2)^2=2^(-(-2)). This problem is the same as finding x for x^2=2^x where 2 and 4 are quite obvious only the third solution is a bit tricky, all just with a negative sign.
That Formation of expression to apply W function was great.
Wtf I didn't understand a shit
@@sherueatyourbestfriend6791 2:39
Here in the first eq. you can't apply W -Function
But he nicely took the square-roots on both sides and multiply by ln(2)/2
That was what I like.
Thank you for your great effort!
My pleasure!
You are really good at explaining that well....go on mahn..
Absolutely beautiful!
Nobody
6th graders watching this video: "INTERESTING..."
I spotted the solution -2, then solved using the Lambert W0 function to get 2 solutions (one positive and the other negative which I assumed to be -2), but alas didn't spot the obvious solution -4 nor did I think of using W-1. So I really appreciated your careful approach!
People actually understand all this stuff is just mindblowing.
I like your funny words magic man
I am a korean and your explanation is easier than learning in korean. So, can you make a video of defining ln and e ? It will be really helpful for me. If there is, i really want the link. Thanks.
ruclips.net/video/Q9puUgDc2BY/видео.html
Nice video! You seemingly ignored the other infinite answers to the equation eg. ~(-7.6545+11.9537i) (from branch 1 of +ve value in W) that come from the lambert w equation, would have been good to at least mention these as the possible complex solutions I think!
Complex solutions are only relevant in engineering. He is tackling it from a pure maths perspective, at least this is my interpretation of his solution.
Excellent video as always!! 😀😀
Thanks a ton!
Thanks for your videos, they help much in my math exam ;)
Glad to hear that!
I tried to solve it using basic algebra, I got one answer -2, this was pretty cool
How did you solve it using algebra ?
I tried super hard but i wasn't able to figure out.
@@ChathuraNuwanga i took log on both sides as my first step and worked ahead and found -2 as solution didnt know it had 3 solutions in it lmao
@@MajimaGoroEnthusiast How did you solve this after this step?
2ln(x) = -x ln(2)
@@ChathuraNuwanga seprate variable and constants in the equation and rewrite it as lnx/x=-ln2/2 apparently you can see x=-2 satisfies the equation
@@MajimaGoroEnthusiast I saw that but -2 does not satisfy ln(x)/x = - (ln2/2) since you can not take log of negative number.
I differentiated both sides to get -2/ln(2)
I loved watching this. Today I learned about the Lambert Funcion, I'm happy I learned something new :D
Glad you enjoyed it!
@@BriTheMathGuy I have replied to your pinned comment, I'd be happy if you could react to it. :)
Hey what are functions ? I really wanna know how to solve the exponent if its x. how should I start the basics ?
Well it’s in this video tho
@@applealvin9167 bro I am in 8th grade so I don't even know what's a function i've tried to learn it but I couldn't thats why I asked what's the basics
@@jasnoor8-d-155 functions are how you turn the shape of lines on a graph into a set of numbers, there are range of function patterns like a simple straight line or a "parabola", to start of with functions you can watch some videos on straight line functions and go from there.
@@itay9458 oh thank you
@@jasnoor8-d-155 a function is simply something that changes the input. If your function f(x) = x + 2 , f(0) gives 2 for example. By the way, with the Lambert W-function you can manipulate functions to make them easier to solve/integrate etc. If you really are in 8th grade by the way, I wouldn't recommend watching videos about the Lambert W function as it is quite advanced stuff. Instead I would just focus on the basics about functions, and all kinds of things you do with them.
in equations like square root of a^x I'd use some sort of property with the absolute values even if the exponent is even or not and get rid of the square root. or other roots if I'm not wrong. I just get the fact that |x|² = x²
edit: this property works for any exponent raised on absolute value so having some correct nth root can further simplify things
I didn't understand a shit
i spoke a bit of nonsense cause i wrote this late at night but this is what i mean
√x² = |x|
∛x³ = |x|
∜x² = √x
Great content. Thanks 😊
Thanks for watching:)
You could use log in the right side, and the -x goes dividing
well
X² = 2^-x .... apply ln to both sides
-> 2 ln ( | x |) = -x ln ( 2 )
--> ln(|x|) / x = -ln(2) / 2
--> ln(|x|) / x = ln(| - 2 |) / (-2)
thus comparatively x = -2
for the positive solution ....... Mr Lambert need your Help please
wait how can both the positive AND negative of ( (x ln 2) / x) work in the lambert function
playback at x1.5 ,,, you're welcome
Me in 8th grade watching this 👀
I thought I could've solved it but I wasn't even close the right answer
We taking positive square root of 1 so it should be 1 and not (+-1).right?
The square root of a value means what number squared equals that value. For example, sqrt(4)= +/- 2 because (2)²=4 and also (-2)²=4. There are two correct answers. There really isn't such a thing as the "positive square root" function, but if you mean the sqrt(+1), well that seems to imply you would expect a negative answer to the sqrt(-1). But that is the definition of i, the imaginary number, and has no bearing on the sqrt(+1).
@@ajreukgjdi94 positive square root function is y=sqrt(x), so if we put x=4 we get corresponding y=2 and not (+-2). Y^2=x is not a function but when we take sqrt in both sides we get 2 branches 1.)y=sqrt(x) which is positive square root and 2.)y=-sqrt(x) which is negative square root.
Take any calculator or ask any learned mathematics teacher sqrt(4) is always written as 2 . However if we have a variable inside the sqrt function (say t^2) then the output is +-t.
Sorry, math noob here. Is W really the best way to express it? I mean, does the inverse function really can't be expressed and simplified using elementary symbols we know?
W is used to solve problems like this, there is indeed no other inverse (in terms of the elementary symbols I think you mean). But really, compare this to using a square root to solve an equation like x^2 = 7. You are happy to say one of the solutions is the square root of 7, even though in some way that doesn't tell you the value either. You can of course express it as a decimal, but that's the same for the W function.
The square root is just more familiar, in some sense more basic, and it has been used for a longer time.
How to be good in maths and develop logical skills ?
omg the Lambert W function strikes again !
Solved using triangle of power from 3b1b
i took log on both sides as my first step and worked ahead and found -2 as solution didnt know it had 3 solutions in it lmao
Ahhh Lambert's W Function. BPRP's favorite
Love the video. Just couldn't ignore the sentence you said at 2:12 "the square root of 1 is +/- 1. Careful! 😏
thanks buddy you made maths easier for me
Glad to hear that!
Solution of this kind of problem depends on constants in it. So there is only constant 2.1/2 is just a variation. Put -2 on both aide aand get 4on both side.
Using newton raphson method i got -2 and 0.76 approximated to 1; using x_1=0 and x_2=1.
Why are you supposing that x²=(1/2)^x ? If I instance x as 3 it doesn't work
We're not saying that this is an *identity*, i.e. that it is true for all values of x.
So, to pick an example of an identity sin 2x = 2 sin(x) cos(x). We can choose any value of x, and the relationship will be true. Or, if you're not familiar with trigonometric functions yet, consider (a + b)(a - b) = a² - b². Again, whatever values of a and b we pick, the identity holds true.
Technically, one should write (a + b)(a - b) ≡ a² - b², and the ≡ symbol (triple bar) makes it clear that it is an identity, not an equation, but in practice the simple = (equals) sign is used.
Identities are useful where we want to change the form of an expression into some other form, that makes it easier to manipulate or solve. We can change the form because we're comfortable in the knowledge that changing the form doesn't change the substance of the expression or change the values of x that make the equation true.
An *equation* on the other hand is usually only true for some specific value(s) of x, or whatever variable it is expressed to be in.
So, if we have, say, an equation 2x + 5 = 23, we can rearrange and solve and get a solution, x = 9. The equation doesn't hold for any other value of x. As equations become more complicated, of higher order (quadratic, cubic etc), or involve trig functions or countless other types of function, equations will have multiple solutions. As a very simple example, x² = 9 has two solutions, which are x = 3 and x = -3. And so on.
Indeed, some equations will have *no* solutions because there are no values which make them valid on both sides. That equation may have been artificially created as a problem question, just to test a student's ability to solve, or it may come about as part of a calculation or proof based on certain assumptions, which are then shown to be impossible. Sometimes there will be no *real* solutions, but delve into the world of complex numbers and we can find solutions that work. These may or may not have some tangible significance in the real world but can exist as mathematical ideas nonetheless.
Hope that helps.
I found two possible solutions to this equation just using some natural logs: x = -2 and x = -4. I'm curious to see what this video shows.
I am not sure how W(y*e^y) = y and W(y) * e^W(y) = y can be derived from the fact that if f(x) = x*e^x then W(x) = f^-1(x)?
Just multiply by 2 on btht side and compare the powers
Great video
But if we go through graph
I am only able to get 2 values of x.
Can anyone explain this??
You’re probably not looking at a large enough graph to also see the intercept at x=-4. Which intercepts are you getting?
@@frogkabobs ya you were right
X will be -4,-2 and .7something
Ah yes , the famous the unknown solution is made of the unknown function x)
is it okay if I don't understand ? (year 2 high school)
lovely so simple i feel dumb i didnt figure out myself
yay math
Ok but what is ln and what's e never heard of these (still at school though)
Equation: has a function in the exponent
My brain: NATURAL LOG NATURAL LOG NATURAL LOGGGGGG
Guys just use log
I would undo both of the initial exponents to get x^(1/x) = +- .5^.5. Take the reciprocal of both sides and we get (1/x)^(1/x) = +- sqrt(2), but from here idk
Hi I am wondering if this math is university level or high school(if so, which class)?
What do you mean high level math? We all learned this in 3rd grade.
I’ve seen the lambert function before but I don’t quite understand it
It is defined simply as the inverse of the function x → xe^x. If that’s not satisfactory, then think about this: How would you define the square root? Well it’s just the inverse of the function x → x². It’s the same idea.
I feel like it could have been easier if you just multiplied both sides by e^(-ln(2)/2) and then it would’ve been xe^x
Or you can guess and check -2 and -4.
As a 10th grader I would stick to hit and trail and stop lol!
e^xln(2) pops out of no where...Is there a video that explains why one can do that... doesn’t one have keep a balance equation?
As f(x)=ln(x) is the inverse function of g(x)=e^x
So ln( e^x ) = x and e^(ln(x)) =x
Practically:
Having 0 apples and multiplying it by itself a total of 0 times gives me 1 apple.
Change my mind.
On my way to mess with my math teacher :))
I felt like this video somehow made me worse at solving.
Tommorow is my Math paper and this is what yt Recommanding me ☠️
revise english
@@bxsim111 plz revise Grammer basic 🤣
By the way Paxtani are telling me to revise English ☠️
And, as expected, those 3 solutions for x we get are the opposite of the results for normal x^2 = 2^x
Which scenario plays out:
0^2 equals 0, or one
(1/2)^0 equals one, 1/2 or 0.
0^2 = 0 and (1/2)^0 = 1
I lost count half video, good video tbh
tf I thought of the first method at the first glance lol
x²=(1/2)^x
x = -∞ or -2
because
(-1/0)²=2^∞
Is using the lambert function the only way to solve this?
It isn’t necessary for the simple solutions (-2,-4), but it’s hard to find a method for finding them that isn’t akin to guess and check. For the third solution, yes, it cannot be expressed in terms of elementary functions, so the lambert W function or some analogue is necessary:
I came here as a beginner and left as an intermediate
ez, graph em
I solved graphically
i paused the video and tried to solve this equation. i took log on both the sides and then i cross multiplied the result and i got lnx/x=-ln2/2 so x=-2
Careful, x=-4 also works
what if x = 0? or x = -2
0² =0, but 2⁰=1. The case x=-2 wa shown in the video.
I particularly liked that part when you didn't explain the step whatsoever and said what you did. Yes, that's half the video
Woww this is really cool
X^2=(-x)^2
My dumbass really read the to the power of x as a square.. shows how i fail just the first step of math
Physics
If you imagine these two equations on graph, you can very easily approximate the solutions.
This seems be the Ramanunjan's way of solving problem , going until infinity 😏
Hi Lucador. If into math competitions , please see
ruclips.net/video/A42ekwVVpHQ/видео.html. We also have a playlist for olympiad 👍
Well, at least I imagine a answer only by look it, obviously it was 0.766 not (-2) 😎👍
I can't tell you how much I hate the Lambert W function - it just feels like the most artificial "cheat" of a function I ever encountered!
Me who already guessed -2 before watching the video
I'm here to see someone prove that 0^0=1
0^0 is considered to be *indeterminate*. Precisely what that word means (and how, for example, it is different to "undefined") is probably, with respect, beyond the level of this question. It's to do with calculus and limits and by all means do some further research into that, but we can look at this in a rather simpler way.
We have two "rules" that we are trying to apply at once, and each rule is giving us a different answer.
One the one hand, we have a rule that says zero raised to any power, is zero. That's not hard to envisage. We can start with zero, multiply it by zero over and over again, and still get zero.
[Side note - when we increase the power we're "multiplying up", when we decrease the power we're "dividing down", so you might see that going from 0^1, which is 0 (anything raised to the power 1 is itself) to 0^0, means dividing by 0, and if you are familiar with that idea you will know it causes its own problems]
On the other hand, we have a rule that says anything raised to the power 0, equals 1. Now that is possibly *less* obvious and you may or may not yet be satisfied that it is the case, but it is a universally accepted rule. Again, you can do a little research on *why* that is the case, and that's a slightly different question.
So when we look at 0^0, we have one rule that says the solution should be 0, and another that says it should be 1. We can't stick to one rule without breaking the other, so we call it indeterminate. It's like being on a road where one sign says "Speed limit 20mph" and a sign next to it says "minimum speed 30mph". We can't have both rules at the same time without breaking one of them.
@@tanelkagan 0 isn't a negative integer, so it wouldn't be dividing.
^0 would read that the base is multiplied by itself a total of 0 times.
And so there should never be a rule suggesting that ^0 should ever equal 1.
Further, 0^0 cannot result in a number greater then 0 when multiplied by a multiple of 0. So even if there were a rule suggesting that x^0 could equal 1, it would not if x=0, because of basic multiplication principles.
I've looked into the reasons why x^0 could equal 1, and they all make less sense then a negative integer having a square root.
It's a mathmatical impossibility.
The question is how are exponents defined?
You multiply the base a number of times indicated by the exponent value by itself.
So in multiplying 0 by itself no times, I'm left with 0.
11th STD maths basics in India
Make me better at solving?
I bet there's gonna be a Lambert W function in this video. Like, it's an algebra equation, and suddenly a Lambert W function appeared out of nowhere. Kinda hate it.
Edit: I was actually correct.
Das Thumbnail zeigt eine Unwahrheit
x^2=2^-x
2ln(|x|)/-x=ln(2)
-ln(|x|)*1/x=ln(2)/2
-ln(|x|)*e^-ln(x)=ln(2)/2
If x>0
-ln(x)=W(ln(2)/2)
x=e^-W(ln(2)/2)
If x
One
x^-1
.
yeʸ
:D
Oh, W(x) function.
So, not real answer.
This outta be interesting.
X=0
2⁰=1, but x²=0
@@frogkabobs explain how 2^0 equals 1 and not two
@@Taime88 any number (aside from 0) to the 0 equals 1. This is because x^(n+1)=x•x^n, and x¹=x. Hence x=x¹=x•x⁰, so x⁰=1.
@@frogkabobs explain how something comes from nothing
@@frogkabobs I have 0 apples. I multiply them by themselves 0 times. I have 1 apple?
Why you didn't write:
x²=2^-x
logx(x²)=log2(2^-x)
2=-x
That means -2=x
The: -2²=2²=4 duh