These videos are extremely helpful especially for this video on comparison test. I understood absolutely everything he said and why he go answers he got. One of the best tutors in the world. Keep it up!!!
Li(x) is just this function with a different lower bound, so it diverges if Li(x) also diverges. Li(x) is also approximately the prime counting function, and since there are infinitely many primes, Li(x) must go to infinity, therefore, this function diverges. QED.
they cant be expressed into 'simpler' form / normal functions we know like sin(x) e^3x etc. (Well technically you can express them via taylor series but they dont really count because the taylor series may not converges for all values)
ʃ 1/ln(x)dx from 2 to x = Lix), which approximates PI(x) the number of primes less than or equal to x as x becomes large ; and I find that interesting. Li is well behaved while PI is not, e.g. PI(9)=PI(10)=4; i.e. 2,3,5,7. ua-cam.com/video/W7YjgmjDpCE/v-deo.html
Will You sine can be integrated without using special functions. Sine, if manipulated in the right way, can become a non-elementary function when integrated. For example, sin(x)/x is non-elementary because we have no way of expressing the integral with normal functions with a defined endpoint to them. Taylor series are an infinite polynomial and are do not terminate. Although they're very useful and (usually) easier to work with, you eventually have to cut off the series and make it an approximation of the real function.
To demonstrate x > lnx, could you have created the function f(x) = x - ln x. And then show that f'(x) is always positive, therefore f(x) is increasing.
a positive derivative of x-lnx still wouldn't be enough proof. You'll need to show that x>ln x for a specific value and due to positive derivative x>ln x for any x greater than that. There are much simpler proofs.
a positive derivative of x-lnx still wouldn't be enough proof. You'll need to show that x>ln x for a specific value and due to positive derivative x>ln x for any x greater than that. There are much simpler proofs.
I remember we using 1/n and (-1)^n/n as basis for comparison tests between sums (discrete or continuous sums, doesn’t matter) so your list is universal lol
If you remember the calculus steps, use the first derivative and set that equal to 0, and see where it’s increasing/decreasing and second derivative where it’s concave up/down. Also, find the find the x- and y-intercepts and the asymptotes. You can verify this on your graphing calculator.
These videos are extremely helpful especially for this video on comparison test. I understood absolutely everything he said and why he go answers he got. One of the best tutors in the world. Keep it up!!!
Thank you very much. You don't know how this video helped me. I understand more with you in English than with my teacher in Portuguese. (Brazil)
Clear presentation of the topics. Thanks sir. DrRahul Rohtak India
Li(x) is just this function with a different lower bound, so it diverges if Li(x) also diverges. Li(x) is also approximately the prime counting function, and since there are infinitely many primes, Li(x) must go to infinity, therefore, this function diverges. QED.
The integral 1/ln(x)dx is li(x) or?
Yea, but that's not elementary
they cant be expressed into 'simpler' form / normal functions we know like sin(x) e^3x etc.
(Well technically you can express them via taylor series but they dont really count because the taylor series may not converges for all values)
ʃ 1/ln(x)dx from 2 to x = Lix), which approximates PI(x) the number of primes less than or equal to x as x becomes large ; and I find that interesting. Li is well behaved while PI is not, e.g. PI(9)=PI(10)=4; i.e. 2,3,5,7.
ua-cam.com/video/W7YjgmjDpCE/v-deo.html
So why is sin(x) a "'simpler' form / normal functions" while li(x) not?
Will You sine can be integrated without using special functions. Sine, if manipulated in the right way, can become a non-elementary function when integrated. For example, sin(x)/x is non-elementary because we have no way of expressing the integral with normal functions with a defined endpoint to them. Taylor series are an infinite polynomial and are do not terminate. Although they're very useful and (usually) easier to work with, you eventually have to cut off the series and make it an approximation of the real function.
I failed my exam last Thursday. But now it seems like I can do all of them with your helping. Thank you.
Is "the List" complete is ln(ln(x))
To demonstrate x > lnx, could you have created the function f(x) = x - ln x. And then show that f'(x) is always positive, therefore f(x) is increasing.
James Dirig 1-1/x isn't always positive
for numbers greater than 1 it is.
a positive derivative of x-lnx still wouldn't be enough proof. You'll need to show that x>ln x for a specific value and due to positive derivative x>ln x for any x greater than that. There are much simpler proofs.
a positive derivative of x-lnx still wouldn't be enough proof. You'll need to show that x>ln x for a specific value and due to positive derivative x>ln x for any x greater than that. There are much simpler proofs.
It is enough. Since f is a growing function, you can say that, for any x greater than 1, f(x)>=f(1)=1>0. Thus, x-ln(x)>0 meaning that x>ln(x).
How do you do this problem with (ln x)^n?
sir u r amazing.. sir Dats clear my confusion... Thnx alot sir from a good Indian😃
ln(x) < x for all positive x.
lim(x→∞)ln(x) = ∞
That's all we need to know for this problem.
I remember we using 1/n and (-1)^n/n as basis for comparison tests between sums (discrete or continuous sums, doesn’t matter) so your list is universal lol
why is it not (1/x^-1)????
For the list you said p has to be positive. But it also has to be greater than 1 right?
That list is just abt the order of magnitude for large n. At that point we don't need to pick p>1
@@gogo-pj2lm But even for large n, ln(x) won't be smaller than x^p if p is e.g. 0.3?
Not really. If u evaluate lim ln(n)/n^0.3, u still can get 0. In fact for any +ve power such lim would still go to 0, and that implies ln(x)
how can i draw graph of 1/lnx
If you remember the calculus steps, use the first derivative and set that equal to 0, and see where it’s increasing/decreasing and second derivative where it’s concave up/down. Also, find the find the x- and y-intercepts and the asymptotes. You can verify this on your graphing calculator.
sir you had made a good
integral 1 to 3 step of π/x
Solutions plz
If you take the substitution lnx=y, you see your integral is divergent
always helpfull !
Hey sir i am from india
My question is
How to intigral of 1/e^tanx dx
Let u be tan x => int 1/e^u•(u^2+1) => e^u
•u + int u/e^u du
Now do integration by parts once more
If ln(1+x) so what
Just sub in u=x+1
Nadie lo explica en Espanol , tengo que parar a ver este video que no entiendo Ni
X2
Hehehe big dx and small dx
but know i need your help if You can. my problem is to find the the value of the next sum :Ln(1) +ln(2)+.....+ln(n)
it's ln(n!)
work
I did not understand haahh