Taylor Series of ln(x) at x = 2, calculus 2 tutorial
HTML-код
- Опубликовано: 2 июн 2016
- Taylor Series of ln(x) at x = 2, problem from James Stewart calculus.
Need to prepare for your calc 2 final? Check out my "100 Calculus 2 problems ultimate review" 👉 • 100 calculus 2 problem...
----------------------------------------
🛍 Shop my math t-shirt & hoodies: amzn.to/3qBeuw6
💪 Get my math notes by becoming a patron: / blackpenredpen
#blackpenredpen #math #calculus #apcalculus
when an asian who speak engslish is more understandable than the proffesor who speak your native language.
good job man
Indians do it better
@@maskbakchodars2271 indians are asians tho
@@marusdod3685 hilarious
@@maskbakchodars2271 yeah no
Wow you made this question so easy, thank you so much! Honestly I couldn't have survived my cal course without this channel.
Thank you so much! I'm a student from Poland and you explained it so clearly that non professor at my university!
Thank you so much! This was explained so much clearer than how my professor explained it!
Great, clear explanation!! Thank you so much! I know I'm gonna do well on my final now! Keep it coming!
THE OOD MIC IS AMAZING!!!!
where is the explanation about finding the radius of convergence?
Thank you for this clear video that has answered most of my questions. Great job.
THANK YOU MAN YOU DONT KNOW HOW YOU HELPED I LOVE YOU SO MUUCHH
You make learning math enjoyable! Thanks so much, I hope you make lots of money through this channel!
THANK YOU!!!!! Im watching all of these!
Thank you kindly! your explanation made it all incredibly clear.
your best work yet!
youre awesome I don't know how i haven't seen your vids before
Excellent explanation! Thanks!
Best video on Taylor series out there, holy
This is a new facial hair combo for me. Very exciting after so many videos!
awesome explanation, thank you!
Thanks so much really helped me understand the concept
Thanks a log ( base e) for beautiful explanation . Dr Taylor Series .
you are amazing and neww beard, that looks really good mashallah brother :)
No one properly explain what (x-2) represent. Is it a shift of the curve by 2 units to the right or we are taking derivative at x=2 on the original graph. Mechanically applying the formula is not important. Contrasting it to expanding (x+dx) is another confusing aspect of Taylor expansion.
so where is the video where u show how to get the r=2?? u said the next vid, but the next vids isnt it. thank you in advance.
I search up t-series on wolfram alpha. Taylor series comes up.
error calculation exercises of taylor plss
Can you please make a video deriving the Taylor series for sqrt( 1 + X ), around X=0. I'm curious how can we derive a pattern for this, seems so difficult. Anyway, very nice video man! I enjoy your videos.
It's just like that: 1+(x/2)-((x^2)/8)+((x^3)/16)-((5x^4)/128)+((7x^5)/256).... and so on, i hope you will see the pattern and continue it yourself. If not, then try do Taylor series for easier functions
awesome!
blackpenredpen. Any chance in the future as to 'why' the Taylor series works, the intuition behind it? How does taking successive derivatives approximate a function? It is not sufficient to learn by rote without understanding the underlying concept. As far as I can discern it has something to do with 'rates of change of the function mirroring the rates of change of the series approximation about a point, and whether that that rate of change is positive or negative. Any lucid explanation would be gratefully appreciated, and thanks.
You could search for 3brown1blue's it has a nice series about this topic
Tudor Cristian. Thanks for your advice, since asking about this I have actually watched 3blue1brown and it has indeed answered my questions regarding this topic. The concept and intuition behind the 'Taylor' series is clearly explained. 3blue1brown and 'professor Leonard's ' series on calculus are far and away the best teaching resource available online, in my opinion. Thanks again.
In same question when we take a=1 then we start 'n' from 1 not from 0 why???
thank you sir !
wouldn't it be better to expand from 1 for obvious reasons?
What happened to the factorial in the denominator? It looks like it became n somehow?
what happened to the factorials?
how to find the 3nd degree of these function xlnx centered at a=1 anyone
untold hero
And the PEANO' REST ?
Thank you
Done
Tks u
sir;
Have you become a beatnik? Man the ‘stache and van dyke do become you. Keep it up. However, you need a black beret to go with it. Love the math!
Why do we have a condition 0 < x
It's called the radius of convergence. It can't converge any farther from the center point (x=a) to the vertical asymptote, on the left of the center point, since there's a discontinuity where it loses all differentiability at the vertical asymptote. From the name "radius of convergence", this implies that the maximum distance on the other side where it converges, it is equal to the radius of convergence. You can prove the radius of convergence works on the right of x=a, with the ratio test for series convergence. Try the ratio test at x=2.01*a, and you'll see that it diverges.
Thanks man
What is Taylor series for ln(abs(x)) centered at -1 ?
Very similar to what it is, when centered at +1, which is the most common example you can look up. You just negate all the odd-ordered terms, to reflect it around the vertical axis.
thank you
Pls Find the points where e^x = tanx
It is probably a stupid question but, isn't Taylor series used for estimations? This expansion just chucks out ln2 + a summation in which if you put x = 2, your whole expression gives you ln2. What is the use of this?
There are some functions where Taylor series have practical applications, like evaluating the erf(x) function. For most functions, there are other series that are much more computationally efficient to evaluate. The main purpose of Taylor series is to give an introductory idea to the concept of using infinite series to approximate transcendental functions, reducing their calculations to just arithmetic and powers, which can be calculated and integrated a lot easier. Taylor series is the easiest infinite series to introduce.
hello, -1^0power is 1? so any negative number to 0 power gives 1? thank you.
TheHouse any number,even complex number, to the zero power is one (only 0^0 is indefined)
You have to snare the (-1), but yes, (-1)^0 is equal to 1. It's everywhere other than 0^0 that is equal to 1.
Solemn explanation for javascript tyranny?
Actually, the ratio test alone would make r=infinity
Please tell me sir summation(£)
£lnx from x=0 to x = infinity
It's value
It diverges, since the sequence diverges.
i almost paid a guy 100$ FOR HIM TO EXPLAIN THIS TO ME thx for saving my grades and pockets
i love u
where did the 2 and 6 in 2[2]^-3/3! and 6[2]^-4/4! go?
It is simplified with the 3! and 4! such that only 3&4 remain at the denominator. It took a while for me to realize that lol
@@ianvideos3149 thanks
i m from india and i must say it was helpful
goat
Aaahh! My name is Adam and I just happened upon this. I feel special! But i'm not........
Wait... isn't this completely pointless because ln 2 is in the actual series?
It's a constant, so it still very easy to work with.
I think you misunderstood. Its not about finding an expression for the taylor series of ln(2) but rather about finding the taylor series of ln(x) *centered* at 2
Instead of 2 I used 1 and got stuck 😅
How old are you,man ?
..what are you say at the begining
adamomo orstovyesky "Adam, this video is for you"
dude youre funny af
he is so similar to tony stark
But your series does not converge onto ln(2).
i love you :v
please find the taylor series for
x^4 . e^(-2x^2) about x=1
Bizim hoca Türkçe konuşmasına rağmen böyle anlatamıyor avvsvsgsgsgs
Could i have your email
14 Idiots