what if we linearly approximate (2.1)^10?

Рет қаралды 10,366

bprp calculus basics

Күн бұрын

Пікірлер: 34

@bprpcalculusbasics 2 жыл бұрын

Try sqrt(8.7) with calculus, NOT calculators! kzbin.info/www/bejne/jKO0d5J6e9OXhZo

@Zeusbeer 2 жыл бұрын

The trick is to write a taylor series, just the quadratic term alone will make the error 16.788

@nikko2505 2 жыл бұрын

Bad idea for 2.1, but 2.01 very well

@emmanuelweiss8672 2 жыл бұрын

"Use differentitial"? :-)

@jamiederinzi 2 жыл бұрын

As an engineer, I would say that within 10% is not a bad approximation 🙂

@phiefer3 2 жыл бұрын

like with all things, error margins are always relative to your specific needs. Plenty of situations where 1536 could be a suitable approximation, and plenty of situations where it's nowhere near accurate enough.

@uggupuggu 2 жыл бұрын

The difference between engineers and people is that engineers go straight to hell

@user-wu8yq1rb9t 2 жыл бұрын

*Calculus, not calculator* Please more of these. I love this category, because I hate calculator, and therefore, at least, for me *calculus not calculator* is such a big relief! Thank you so much dear Teacher 💖

@anshumanagrawal346 2 жыл бұрын

Interesting... As soon as I saw the thumbnail and the video was starting I instinctively factored out the 2 to make 2^10 (1+0.05)^10 because It's been drilled in my head apply binomial like approximation in the (1+x)^n form only

@NoNameAtAll2 2 жыл бұрын

me three

@leickrobinson5186 2 жыл бұрын

NO! Apologies, but this is wrong! It’s *not* necessary for the magnitude of f’(a) to be small. Instead, you need for f’(x) to be *slowly varying* over the interval [a,a+dx]. Counter-example to demonstrate this: The linear function f(x) = 10,000,000 x has a rather large first derivative. However, the local linear approximation for this function is quite good! (perfect, in fact!) :-D (The problem here is that, over the interval from 2 to 2.1, f’(x) changes by more than 50%!)

@Jim-be8sj 2 жыл бұрын

True. This is a good point. Bounds on the error for a linear approximation depend on the magnitude of the second derivative.

@awfuldynne 2 жыл бұрын

just calculus responded to someone else with a similar objection, and you're right. They said the actual requirement is for f'(x) to be approximately constant across the dx interval.

@NoNameAtAll2 2 жыл бұрын

what about f(x) = 10,000,000x + sin(10,000,000x)? f' would change drastically, but significance of answer isn't really affected other than close to x=0

@leickrobinson5186 2 жыл бұрын

@@NoNameAtAll2 Yes, that would be another example of what I am saying.

@NoNameAtAll2 2 жыл бұрын

@@leickrobinson5186 you are saying "f'(x) needs to be slowly varying" I posted function that has f' variating wildly fast

@Engy_Wuck 11 ай бұрын

if the approximation is "good enough" depends on what you want. If you only want to know is "will 2.1^10 be bigger than 2000", then the approximation works. "Is it nearly 1500" - well, not so much. As quite often it depends on what you want answered.

@edskev7696 2 жыл бұрын

I disagree that f' needs to be small for the linear approximation to work. Rather, you need it to approximately constant over the region of interest. That is, to approximate f(x) using values at a, then f' needs to be approximately constant between a and x.

@bprpcalculusbasics 2 жыл бұрын

I think we have the same idea. When I said f' should be small, I was implying that f shouldn't be changing that much around there.

@TheAmazingFocus 2 жыл бұрын

@@bprpcalculusbasics I think it's more accurate to say the second derivative of f should be small. Taylor's theorem gives an upper bound on the error, and a first-order Taylor series has an error of f''(c)*(x-a)^2/2! for some appropriate c. If f(x)=(2+x)^10, then the formula shows the error is between 115.2 and 170.2, which is the case!

@nurrohmadi7852 2 жыл бұрын

why not use exponent method? (21/10)^10 ((21/10)^2)^5 (((21/10)^2)^3)•(((21/10)^2)^2) work a litle bit and we get aproxtimation and you can stop if get dots