/sci/ - Science & Math » Thread #11040099

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Anonymous Mon 07 Oct 20:02:35 2019 No.11040099 View Reply Original Report

Quoted By: >>11040104 >>11040161 >>11040200 >>11040211 >>11040237 >>11040241 >>11040246 >>11040870 >>11041551

> he didn't see this proof in his highschool calculs course
> He took this limit for granted

Laughs in superior Africain education

Anonymous

Anonymous Mon 07 Oct 2019 20:04:13 No.11040104 Report

Quoted By: >>11040122 >>11040763

>>11040099
>Calling a drawing a proof

Anonymous

Anonymous Mon 07 Oct 2019 20:10:21 No.11040122 Report

Quoted By:

>>11040104
It is to highschoolers

Anonymous

Anonymous Mon 07 Oct 2019 20:23:27 No.11040161 Report

Quoted By: >>11040222

>>11040099
An easier approach without using squeeze is just approximating arc length.
$\lim\limits_{n \to \infty} n\sin(\theta/n) = \theta$

Anonymous

Anonymous Mon 07 Oct 2019 20:39:19 No.11040200 Report

Quoted By: >>11040223

>>11040099
>using geometry in an analysis proof
You can't do that.

Anonymous

Anonymous Mon 07 Oct 2019 20:42:35 No.11040211 Report

Quoted By: >>11040222

>>11040099
>Africain
>site name written in portuguese
Angolano ou de Cabo Verde?

Anonymous

Anonymous Mon 07 Oct 2019 20:44:35 No.11040222 Report

Quoted By:

>>11040211
Just an image i copied from google
>>11040161
Elaborate ?

Anonymous

Anonymous Mon 07 Oct 2019 20:44:54 No.11040223 Report

Quoted By:

>>11040200
sure you can

Anonymous

Anonymous Mon 07 Oct 2019 20:48:34 No.11040237 Report

Quoted By: >>11040258 >>11040785

>>11040099
>highschool calculs course
In 13th grade we had Taylor expansions. $\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} + \dots$

Anonymous

Anonymous Mon 07 Oct 2019 20:50:11 No.11040241 Report

Quoted By:

>>11040099
This limit is intuitively obvious

Anonymous

Anonymous Mon 07 Oct 2019 20:52:03 No.11040246 Report

Quoted By: >>11040250 >>11040255 >>11040889

>>11040099
We learned it via the squeeze theorem and it was a lot more rigorous

Anonymous

Anonymous Mon 07 Oct 2019 20:53:06 No.11040250 Report

Quoted By: >>11040889

>>11040246
the picture is literally squeeze theorem

Anonymous

Anonymous Mon 07 Oct 2019 20:53:57 No.11040255 Report

Quoted By: >>11040258 >>11040889

>>11040246
Wait I'm stupid, the pic is basically just the squeeze theorem. My education was still a lot more rigorous.

Anonymous

Anonymous Mon 07 Oct 2019 20:56:32 No.11040258 Report

Quoted By: >>11040264 >>11041845

>>11040255
This pic just skips a few steps
>>11040237
> 13th grade
What ?

Anonymous

Anonymous Mon 07 Oct 2019 20:59:19 No.11040264 Report

Quoted By:

>>11040258
We start counting at 1=first grade, lol

Anonymous

Anonymous Tue 08 Oct 2019 00:29:34 No.11040763 Report

Quoted By:

>>11040104
...my Topology Prof used to call it "proof by picture", and would show us examples in class from various Math journals.
Nothing wrong with OP diagram except the omission of AB=1

Anonymous

Anonymous Tue 08 Oct 2019 00:41:24 No.11040785 Report

Quoted By: >>11041502 >>11041835

You haven't proven anything until you rigorously define your notion of area. Why is it that a shape which is contained in another shape has a smaller area? Come on, now.
>>11040237
You need to use the limit you're proving to find the taylor expansion, moron.

Anonymous

Anonymous Tue 08 Oct 2019 00:52:21 No.11040814 Report

Quoted By:

No, I didn't. The proof I studied actually made sense geometrically.

Anonymous

Anonymous Tue 08 Oct 2019 01:16:20 No.11040870 Report

Quoted By: >>11040874 >>11040880

>>11040099
sin x <= x is "obvious"
x cos x <= x is "not obvious"

Anonymous

Anonymous Tue 08 Oct 2019 01:17:23 No.11040874 Report

Quoted By: >>11040901 >>11041303

>>11040870
i meant
x cos x <= sin x is "not obvious"

Anonymous

Anonymous Tue 08 Oct 2019 01:19:04 No.11040880 Report

Quoted By:

>>11040870
>x cos x <= x is "not obvious"
sure it is. for any x<<1, x will be some number greater than zero and cos x will be some number slightly less than 1. a number slightly less than 1 multiplied by x will always be less than x.

Anonymous

Anonymous Tue 08 Oct 2019 01:26:51 No.11040889 Report

Quoted By:

>>11040246
>>11040250
>>11040255
These three posts are /sci/ in a nutshell

Anonymous

Anonymous Tue 08 Oct 2019 01:34:37 No.11040901 Report

Quoted By: >>11041343

>>11040874
bump for someone smarter than me to give an intuitive reason why t*cos t<sin t

Anonymous

Anonymous Tue 08 Oct 2019 04:56:40 No.11041303 Report

Quoted By:

>>11040874
why the quotes?

Anonymous

Anonymous Tue 08 Oct 2019 05:21:31 No.11041343 Report

Quoted By:

>>11040901
It's not even true
If it was, then t < tan t
Which is false

Anonymous

Anonymous Tue 08 Oct 2019 05:38:48 No.11041375 Report

Quoted By: >>11041398 >>11041680

Just use l'Hôpital lmao

Anonymous

Anonymous Tue 08 Oct 2019 05:46:13 No.11041398 Report

Quoted By:

>>11041375
based and engineering pilled

Anonymous

Anonymous Tue 08 Oct 2019 06:41:35 No.11041473 Report

Quoted By:

just... look at the y = sin(x) graph and the y = x graph and see that theyre equivalent around 0 so its like saying x/x and 0.0000001/0.0000001 is... 1!

Anonymous

Anonymous Tue 08 Oct 2019 07:00:44 No.11041502 Report

Quoted By:

>>11040785
>You need to use the limit you're proving to find the taylor expansion, moron.
not if the series is his definition of sin(x)

Anonymous

Anonymous Tue 08 Oct 2019 07:39:02 No.11041551 Report

Quoted By:

>>11040099
Tbh Im frm a developing cuntry and we didn’t learn calculus in highschool till college.
Youre lucky anon.

Anonymous

Anonymous Tue 08 Oct 2019 09:47:27 No.11041680 Report

Quoted By:

>>11041375
You need this limit to determine the derivative which you need for l'hôpital

Anonymous

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Anonymous Tue 08 Oct 2019 11:37:07 No.11041835 Report

Quoted By:

>>11040785
>You need to use the limit you're proving to find the taylor expansion, moron.
No, you need to find derivatives at x=0 to find the taylor expansion, moron. Inb4 that one case where a taylor expansion fails at x=0. If you want to constrain your error below 1%, you can use $\sin(\theta) \approx \theta$ up to $\theta = 0.245 \approx 14^{\circ}$ .

Anonymous

Anonymous Tue 08 Oct 2019 11:41:21 No.11041845 Report

Quoted By: >>11041850

>>11040258
Germany used to have 13 grades up until a few years ago, and still has for some school forms in some states.

Anonymous

Anonymous Tue 08 Oct 2019 11:45:40 No.11041850 Report

Quoted By:

>>11041845
They have reintroduced it now in some federated states because they saw that 12 years at the same level of education was not possible.

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