Professor Lewis thank you for this classic selection of functions that is used to find Taylor Series Expansion in Calculus II. Taylor Series Expansion is a powerful method in science and engineering.
Here's the course page on the Taylor Series: ocw.mit.edu/courses/18-01sc-single-variable-calculus-fall-2010/pages/unit-5-exploring-the-infinite/part-b-taylor-series/. Best wishes on your studies!
Good to memorize the easiest Maclaurin series off of Wikipedia; - sin, cos, sinh, cosh, arctan, arctanh - e^x, ln(1-x), ln(1+x) - 1/(1-x), 1/(1-x)^2, 1/(1-x)^3
I have the same question, i thought it's because he add the two series when the n is even, and for that he had 2(Serie), and that 2 cancel out with the 1/2, but i don't know ir thats correct
I hope my calculus processor will be a good processor because I don't want a crappy processor while in college. *DONT TELL ME PROCESSOR IS NOT THE RIGHT WORD!*
Hey Karthik! Consider the nth term here: t_n = (-1)^n ( (2x)^(2n+1))/(2n+1)! Observe that the index runs from 0 to infinity Consider 1)The first term n = 0 Term1 = (-1)^0 ( (2x)^(1))/1! = 2x 2) Second term n = 1 Term2 = (-1)^1 ( (2x)^(2+1))/(2+1)! = -(2x)^3/3! 3) Third term n = 2 Term3 = (-1)^2 ( (2x)^(4+1))/(4+1)! = (2x)^5/5! The sum of all such tems would give you the series expansion. If I understand the motive behind your question properly, notice that it's (-1)^n and the power of x is (2n+1). See that (-1)^(2n+1) would not alter the signs but (-1)^n would alter the signs of odd terms as the index runs from 0 to infinity.
Professor Lewis thank you for this classic selection of functions that is used to find Taylor Series Expansion in Calculus II. Taylor Series Expansion is a powerful method in science and engineering.
Taylor series is basically McLaurin expanding around a different point. If you set your expansion point to zero, then you get the McLaurin series.
smooth explanations, if our professors where like that, i would be happy, but they are some elders stack in the 60s
at 7:02 he says we've already seen whatìs the taylor series for ln functions. anyone knows where i can find that lecture?
Here's the course page on the Taylor Series: ocw.mit.edu/courses/18-01sc-single-variable-calculus-fall-2010/pages/unit-5-exploring-the-infinite/part-b-taylor-series/. Best wishes on your studies!
this is like going back to school... great explanation!!
? Almost all other Unis do that too.
If you do Electrical Engineering, you tend to get them really early on with complex analysis.
and if you take the IB Diploma you do it during your senior years in high school for your option topic
Madhava serie not Taylor serie
Nice explanations and videos!
it is a very good explanation, very good video
Good to memorize the easiest Maclaurin series off of Wikipedia;
- sin, cos, sinh, cosh, arctan, arctanh
- e^x, ln(1-x), ln(1+x)
- 1/(1-x), 1/(1-x)^2, 1/(1-x)^3
I Really Like The Video Finding Taylor's Series From Your
Where is the 1/2 in the front go? I see 1/2 in front of the original two infinite sums, but thereafter it disappears. Where did it go?
I have the same question, i thought it's because he add the two series when the n is even, and for that he had 2(Serie), and that 2 cancel out with the 1/2, but i don't know ir thats correct
when n is even -1 goes away and we'll get 2x^n at the numerator of the sum of the two series.
I love this dude
I hope my calculus processor will be a good processor because I don't want a crappy processor while in college. *DONT TELL ME PROCESSOR IS NOT THE RIGHT WORD!*
that's a huge piece of chalk.
crystal clear
It's like he's using sidewalk chalk.
hey does anyone know who came up with the multivariate expression for the taylor expansion, you know the one with the iterated partial derivatives
+Adam Ledger I'm pretty sure it was recent but the mind plays tricks from time to time
Ayuda, como resuelvo: integral de x / (raiz(x) + 2)
how did i go from learning logs to this? its good because i do need 2 know infint geometric serieses but wow
Why arent there factoriels in ln function?
how can we use (-1)^n?
even odds are alternating + & -
(at 6.06)
Hey Karthik!
Consider the nth term here:
t_n = (-1)^n ( (2x)^(2n+1))/(2n+1)!
Observe that the index runs from 0 to infinity
Consider
1)The first term
n = 0
Term1 = (-1)^0 ( (2x)^(1))/1! = 2x
2) Second term
n = 1
Term2 = (-1)^1 ( (2x)^(2+1))/(2+1)! = -(2x)^3/3!
3) Third term
n = 2
Term3 = (-1)^2 ( (2x)^(4+1))/(4+1)! = (2x)^5/5!
The sum of all such tems would give you the series expansion.
If I understand the motive behind your question properly, notice that it's (-1)^n and the power of x is (2n+1). See that (-1)^(2n+1) would not alter the signs but (-1)^n would alter the signs of odd terms as the index runs from 0 to infinity.
I'm interested if they teach how to solve the problems given in IMO (International Mathematical Olympiad) in MIT. Could anyone answer, please?
Katy Lee I think not. Only those who have not attended university are eligible to compete in IMO.
Excelent, useful to study
what happened to his voice at 3:55???
awesome boss..
@brycepatties I bet you say that to all your professors!
Very helpful
Multiple is down simple.
Being very polite non of this is used in real life
Esto buscaba 🙋
i learnt trigonometry today
now by brain hurts
MIT does taylor series in calc 1? wtf!
I love ya, you’re a great professor, but you remind me too much of Jon Arbuckle.