the simplest approach to power series.

[@logician1234]

Nice, I'm surprised this approach isn't used more often in teaching

[@sochjan]

The answer is very simple. Integration is simply after power series in Calculus.😅

[@NintendoGamer789]

@@sochjan Bruh I had integration covered as the first topic, power series was at the end (for calc 2)

[@karlmudsam2834]

Calculus at my school is pretty general, a lot of different students take it, and that means less rigor, which is a shame, cause you miss out on things like this, but I’d argue less rigor is a good thing at that level

[@normanstevens4924]

If you're doing a rigorous analysis course then you'll have defined the exponential function and sin and cos in terms of power series anyway.

[@logician1234]

@@normanstevens4924 I always hated this definition, it's literally backwards

[@paulgillespie542]

Interesting. This provides excellent insight into Taylor series ideas before the introduction of more formal definitions. Well done Professor!

[@MathFromAlphaToOmega]

If I recall correctly, the proof of the error bound on the remainder for Taylor series actually uses something like this. If f is a function whose nth Taylor polynomial is P_n(x), then the first n derivatives of P_n match those of f at x=0, and the (n+1)st derivative of P_n is 0. So if we let R_n(x)=f(x)-P_n(x) be the error, we find that the (n+1)st derivative of R_n equals that of f. Then we take M to be the max of |f^(n+1)| on the interval [0,a], and then that's also the max of the (n+1)st derivative of R_n. Integrating n+1 times from 0 to x gives |R_n(x)|≤Mx^(n+1)/(n+1)! for 0≤x≤a.

[@goodplacetostop2973]

17:34

[@zlodevil426]

Good place to st

[@atzuras]

brilliant. also it works to demonstrate sin x < x and lim x->0 (1- cos x)/x =0

[@martincohen8991]

I first saw this in Dorrie's "100 Great Problems of Elementary Mathematics", which is highly recommended.

[@wilderuhl3450]

“And that’s a good place to st-“ Dang, he really was eager to stop.

[@axelperezmachado3500]

I mean... it was a pretty good place

[@GlenMacDonald]

[9:34] Suggestion when doing the induction: assume BOTH equalities are true for n = k, and prove BOTH equalities are true for n = k +1. This avoids the somewhat awkward idea of a "ladder".

[@DanielDH179]

7:55 k should be from 0 to m.

[@orenfivel6247]

with combined similar approaches of sin, cos (parity approach) and exp (monotone inc appr.), we can obtain power series for cosh and sinh easily

[@ThAlEdison]

I may have missed it, but I would've liked an explanation of why you started with cos(t)

[@Jack_Callcott_AU]

At 7:45 there is a mistake: m, not infinity, should be the upper bound of summation.

[@krisbrandenberger544]

No, m should be the upper bound for that sum and the one @ 8:05.

[@jamesmstern]

That's beautiful.

[@mathechne]

very clear!

[@Alan-zf2tt]

I wonder if a visual aid might help? Displaying those inequalities as graphs - well it is easy-peasy now - might suggest a convincing argument (eg proof?) is required for all naysayers? Or even differentiating power series to show - well what it shows and how it links in nicely with integration rules?

[@MathCuriousity]

Can somebody xplain right off the bat and why it’s always true that that inequalities hold if we integrate both sides of an inequality ? I’d like to look up the theorem or concept!!

[@paulgillespie542]

Agreed. However, this insight would have helped my initial understanding of Taylor series on a gut level. Wrt DrROBERT's comment.

[@CM63_France]

Hi, I like this sort of proof "at lower cost". 7:57 : up to m and not to infinity 17:35 : missing "op".

[@emanuellandeholm5657]

Don't you also need something like x < 1? cos x and sin x are bounded, but any finite polynomial will eventually be arbitrarily large for large enough x.

[@angel-ig]

But not every term in the series is positive. It works for all x

[@emanuellandeholm5657]

@@angel-ig How can that be? For x large enough, the highest power will just dominate, and that term will grow without limit as x tends to infinity. Edit: Wait, the truncated series will grow indefinitely, but in the right direction, I see that now. You get -oo

[@angel-ig]

@@emanuellandeholm5657 Yeah, the truncated polynomials are unbounded but the limit series is equal to cos(x) or sin(x) for all x

[@MathCuriousity]

Why do his arrows look like phallic symbols?!

[@CousinoMacul]

The "fact" in your video is a form of the Cauchy-Schwarz inequality. 😊

[@erfanmohagheghian707]

Just wondering how on earth you would call this the simplest approach! Lol

[@mollejalopez8012]

Me encanta como lo explicas y lo que haces. Cásate conmigo ❤❤

[@CM63_France]

Ya esta casado, con una niña.

[@mollejalopez8012]

​@@CM63_France😢

[@DrR0BERT]

I would not consider this a simple approach, let alone the "Simplest".

[@blackcat5771]

Cool