Thanks for your teaching 👌

Thanks respected sir, love from India

Perfect 👌🏼

@onlineMathsTV

Жыл бұрын

Thanks for watching our videos and dropping this encouraging comment ma. We all @OnlinemathsTV love you dearly ma 💖💖❤️❤️💕💕

Thanks teacher, good equation, I like math genius

In the original equation, you have lnx, thus x must be positive. As such, the negative solution should be rejected

x^2=-5lnx raise e to both sides e^x^2=(e^lnx)^-5=x^-5 raise both sides to 1/x^2 power e=x^-5/x^2 raise both sides to -1/5 power e^-1/5=x^1/x^2 im gonna try to use my proof for a similar problem with x^1/x instead of 1/x^2 to help get a picture on how to solve this let x^(1/x^2)=y x^(1/x^2)=y y^x^2=x lny^x^2=lnx (x^2)lny=lnx divide both sides by x^2 lny=(lnx)/x^2=(1/x^2)lnx=(x^-2)lnx x^-2=e^ln(x^-2)=e^-2lnx =(lnx)(e^-2lnx)=lny *-2 both sides -2lnx*e^-2lnx=-2lny -2lnx=W(-2lny) -2lnx=lnx^-2 raise e both sides e^-2lnx=x^-2 x^-2=1/x^2=e^W(-2lny) x^2=1/e^W(-2lny)=e^-W(-2lny) x=sqrt(e^-W(-2lny)) if x^(1/x^2)=y, x=sqrt(e^-W(-2lny) overall conclusion: if x^(x^n)=y x=(e^W(nlny))^n => if x^(1/x^2)=e^-1/5 x=sqrt(e^-W(-2ln(e^-1/5))) =sqrt(e^-W(2/5))

Great! Congrats sir. x = ± √(e^-W(2/5) another form is *x = ± √[(5/2)W(2/5)]* because e^W(x) = x/W(x) what comes from W(x)e^W(x) = x But I believe the negative value should be rejected because of ln(x) in the equation.

very nice question

Wow! This is a very nice solution. Well done Jakes....👍👍👍

@onlineMathsTV

Жыл бұрын

Thanks a million @Danniel Franca for watching our content and at the same time commenting. Much love ma...❤️❤️💕💕💕

I like your mathematical level of explanation bro, thanks for this ability and thanks for making me understand the Lambert W function in a simple way. Mor grace from Above son.

@onlineMathsTV

Жыл бұрын

Much appreciated sir and thanks for watch our contents

Respect 🙏🙏🙏

@onlineMathsTV

Жыл бұрын

Thanks for appreciating our little effort. Also thanks for watching and dropping this encouraging comment. Much love from everyone of us @onlinemathsTV 💕💕❤️❤️💖💖

great

@onlineMathsTV

Жыл бұрын

Thanks for watching and leaving this comment sir. Much love from everyone @onlinemathsTV 💕💕💕

Nice, but remember, since we have a logarithm, the argument must be positive, so x>0, we have to exclude the negative solution :)

@onlineMathsTV

Жыл бұрын

Yap...thanks a million sir.

@gracyledsy356

Жыл бұрын

What grade his teaching?

@kylekatarn1986

Жыл бұрын

@@gracyledsy356 you mean, mine? High School from Italy

@gracyledsy356

Жыл бұрын

@@kylekatarn1986 nice to meet u, I mean this video his teaching for what grade?

@kylekatarn1986

Жыл бұрын

@@gracyledsy356 I think this is for challenges, like the ones you should find during the Olimpyades

Thank you for introducing me to the Lambert W function. I am a retired former lecturer in physics to university students and have considerable interest in mathematical aspects of physics but, perhaps to my shame, confess that I have not met the Lambert function before- I don't think it appears in many of the well-known text books ( e.g. the widely recommended book by G Arfken) on mathematical methods for physics or similar books. You have given me a motivation to learn something about this function from online presentations. By introducing the function f(x) = xsquared + 5 ln x and then finding the root of this function [i.e. setting f(x)=0] using the Newton-Raphson method of successive approximations I find the answer x=0.86193...... As you admit yourself, you did not finish with a numerical value for x but instead gave the result in terms involving the Lambert function. I find it puzzling that your final result for x in this presentation was +/- a particular value because if we agree that x=0.86193... is indeed one solution then how can x=- 0.86193... be another solution given that this would mean that the right-hand side of the original equation would then be complex but xsquared on the left hand side would still be real? The comment by @kylekatarn1986 essentially makes the same point.

@onlineMathsTV

Жыл бұрын

So sorry the reply to this comment of yours is coming a bit late sir. Your answer is correct from the Newton-Raphson method you applied. As for me, I make use of the WolframAlpha calculator to get my numerical values from this point/stage of the math. As for the minus sign, it will give positive answer also after the square root evaluation and so it will come out as 0.86193....because (-)×(-)=+. Remember we took the square root of both sides of the equation @9:55 Thanks so much for watching and at the same time dropping this wonderful comment sir. Much love and respect ❤️❤️💖💕💕🙏🙏🙏

Now the only thing to do is to run a computer program to approximate the numerical value of x, since the "W" function is only a definition...but of course we may have special purpose calculators to do the...

@onlineMathsTV

Жыл бұрын

👍👍👌👌👌

🙏🙏🙏

@onlineMathsTV

Жыл бұрын

Thanks a million for watching our video tutorial and dropping this wonderful symbol of encouragement ma. We all @onlinemathstv love you dearly ❤️❤️💖💖💕💕💕

How do you get a numerical value?

@keithturvey5818

Жыл бұрын

@hokie6384 One way to get a numerical value for x is to use a technique of successive approximations, e.g the Newton -Raphson method. I obtain x=0.86193.... although one should first find an approximate result before applying that method. In my case I initially found that x=0.5 and x=1.0 lie on opposite sides of the correct answer. For anyone adept at writing computer programs to numerically solve equations ( which I am not ) it would be an easy task to write a suitable successive approximations program. You may find my comment elsewhere amongst the responses of interest. I struggled a bit with Mr Jakes accent but I think he indicated that in a future online presentation he would show us how to get a numerical result- perhaps it will be more elegant than my way.

@onlineMathsTV

Жыл бұрын

Use the WolframAlpha calculator sir.

Not clear to me

Yanlış 2/5 değil -2/5 olmalı

@onlineMathsTV

9 ай бұрын

Sorry I did not get your point here sir, kindly point out the error again in a clearer form. Thanks sir.

Sorry, x can't be negative

@onlineMathsTV

5 ай бұрын

ok

False solve