/sci/ - Science & Math » Thread #13529013

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Anonymous Tue 17 Aug 15:35:38 2021 No.13529013 View Reply Original Report

Quoted By: >>13529059 >>13529107 >>13529120 >>13529210 >>13529215 >>13530331 >>13531664

How does the third inequality follow from the first?

Anonymous

Anonymous Tue 17 Aug 2021 15:39:33 No.13529032 Report

Quoted By: >>13529120

Idk. No one on this board actually knows any math

Anonymous

Anonymous Tue 17 Aug 2021 15:46:01 No.13529059 Report

Quoted By:

>>13529013
Forgot to mention: |L|=/=0

Anonymous

Anonymous Tue 17 Aug 2021 15:54:13 No.13529092 Report

Quoted By:

I've been trying to find how for 40 minutes and I still can't see it.

Anonymous

Anonymous Tue 17 Aug 2021 15:57:40 No.13529101 Report

Quoted By: >>13529128

it's literally the triangle inequality??

Anonymous

Anonymous Tue 17 Aug 2021 15:58:54 No.13529107 Report

Quoted By:

>>13529013
Take the non-standard analysis pill goy

Anonymous

Anonymous Tue 17 Aug 2021 16:00:02 No.13529109 Report

Quoted By:

Because math is made up bullshit so you can make up any bullshit you want and call it math

Anonymous

Anonymous Tue 17 Aug 2021 16:02:29 No.13529120 Report

Quoted By: >>13529142

>>13529032
Silence brainlet.
>>13529013
|L| <= |f(x)| + |f(x) - L| (triangle inequality)
move |f(x)- L| to the other side,
so |f(x)| >= |L| - |f(x) - L| >= |L| - |L|/2 = |L|/2

Anonymous

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Anonymous Tue 17 Aug 2021 16:03:50 No.13529128 Report

Quoted By: >>13529135

>>13529101
How?
Also i don't get why the last inequality is closed instead of open.
|M|/(|M|+1)<1 for all |M|

Anonymous

Anonymous Tue 17 Aug 2021 16:04:51 No.13529135 Report

Quoted By:

>>13529128
*last inequality for this image

Anonymous

Anonymous Tue 17 Aug 2021 16:05:39 No.13529137 Report

Quoted By: >>13529167

From the properties of the absolute value:
$\left|f\left(x\right)\,-\,L\right| \;\geq\; \left|L\right|\,-\,\left|f\left(x\right)\right|$
Therofore:
$\left|L\right|\,-\,\left|f\left(x\right)\right| \;<\; \frac{\left|L\right|}2$
Add $\left|f\left(x\right)\right|\,-\,\left|L\right|\,/\,2$ on both sides.

Anonymous

Anonymous Tue 17 Aug 2021 16:06:04 No.13529140 Report

Quoted By: >>13529174

Is there fore to it? Because that last line implies that f(x) can be arbitrarily larger than L, which contradicts the first line?

Anonymous

Anonymous Tue 17 Aug 2021 16:06:22 No.13529142 Report

Quoted By: >>13529192

>>13529120
>|L| - |f(x) - L| >= |L| - |L|/2
Shouldn't it be |L| - |f(x) - L| > |L| - |L|/2
Because |L|/2 > |f(x) - L|

Anonymous

Anonymous Tue 17 Aug 2021 16:13:49 No.13529167 Report

Quoted By: >>13529202

>>13529137
But how do you get the closed the interval sign like in the OP image

Anonymous

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Anonymous Tue 17 Aug 2021 16:15:00 No.13529174 Report

Quoted By:

>>13529140
This is the entire theorem and proof

Anonymous

Anonymous Tue 17 Aug 2021 16:19:21 No.13529192 Report

Quoted By: >>13529195

>>13529142
Yes.

Anonymous

Anonymous Tue 17 Aug 2021 16:19:45 No.13529193 Report

Quoted By:

looks like some people are getting filtered here

Anonymous

Anonymous Tue 17 Aug 2021 16:20:21 No.13529195 Report

Quoted By: >>13529228

>>13529192
So the book is wrong?

Anonymous

Anonymous Tue 17 Aug 2021 16:23:02 No.13529202 Report

Quoted By: >>13529222 >>13530317

>>13529167
$>$ implies $\geq$ .

Anonymous

Anonymous Tue 17 Aug 2021 16:26:38 No.13529210 Report

Quoted By: >>13529219

>>13529013
|f - L| < |L|/2
add |-f|
|-f| + |f-L| < |L|/2 + |-f|
triangle lhs
|-L| <= |-f| + |f-L| < |L|/2 + |-f|

Anonymous

Anonymous Tue 17 Aug 2021 16:29:25 No.13529215 Report

Quoted By: >>13529224

>>13529013
Ignore the autistic replies. For x close enough to x_0, f(x) is close to L, so you can arrange that the distance from f(x) to L is smaller than |L|/2 (which is a positive number) for all x close enough to x_0.

Anonymous

Anonymous Tue 17 Aug 2021 16:29:42 No.13529219 Report

Quoted By: >>13530317

>>13529210
But there should be <= instead of just <

Anonymous

Anonymous Tue 17 Aug 2021 16:30:13 No.13529221 Report

Quoted By: >>13529230

reverse triangle ineq:

||f|-|L|| <= |f-L|<|L|/2
So |f|-|L|>-|L|/2

Anonymous

Anonymous Tue 17 Aug 2021 16:30:44 No.13529222 Report

Quoted By: >>13529266 >>13531775

>>13529202
How?
X>0 is not the same as X=>0

Anonymous

Anonymous Tue 17 Aug 2021 16:31:47 No.13529224 Report

Quoted By: >>13529231

>>13529215
I get the first inequality, but how does the third follow? I'm thinking the book is wrong in making it a closed interval sign

Anonymous

Anonymous Tue 17 Aug 2021 16:32:19 No.13529228 Report

Quoted By:

>>13529195
No.

Anonymous

Anonymous Tue 17 Aug 2021 16:33:12 No.13529230 Report

Quoted By: >>13529235

>>13529221
Exactly, I think the book is wrong in putting => instead of just > right?

Anonymous

Anonymous Tue 17 Aug 2021 16:33:35 No.13529231 Report

Quoted By:

>>13529224
If f(x) is close to L, then |f(x)| is close to |L|, so naturally |f(x)| > |L|/2 for x close enough to x_0

OP Tue 17 Aug 2021 16:34:38 No.13529234 Report

Quoted By: >>13529242

Just to be clear I think it should be:
|f(x)|>|L|-|L|/2
Instead of:
|f(x)|>=|L|-|L|/2

Anonymous

Anonymous Tue 17 Aug 2021 16:35:10 No.13529235 Report

Quoted By: >>13529241

>>13529230
It's not wrong, just weaker

OP Tue 17 Aug 2021 16:37:41 No.13529241 Report

Quoted By: >>13529247

>>13529235
Why isn't it wrong? For example:
f(x)=1/x for x>=0 is wrong because f(0) is undefined
But f(x)=1/x for x>0 is fine

Anonymous

Anonymous Tue 17 Aug 2021 16:37:45 No.13529242 Report

Quoted By: >>13529249

>>13529234
The first inequality is stronger than the second one but both hold. Textbooks are not too economical with the estimates used sometimes.

Anonymous

Anonymous Tue 17 Aug 2021 16:39:44 No.13529247 Report

Quoted By:

>>13529241
You make no sense. > implies =>. => is weaker

OP Tue 17 Aug 2021 16:40:22 No.13529249 Report

Quoted By: >>13529269

>>13529242
Could you explain what you mean by "stronger"?

Anonymous

Anonymous Tue 17 Aug 2021 16:45:13 No.13529266 Report

Quoted By: >>13529279

>>13529222
"Implies" doesn't mean "is the same as".
If A is greater than B, of course A is greater than or equal to B.

Anonymous

Anonymous Tue 17 Aug 2021 16:46:00 No.13529269 Report

Quoted By: >>13529279 >>13529341

>>13529249
The first inequality implies the second inequality, so it's stronger. For example, 2 > 1 and 2 >= 1 both hold. The first inequality holds for obvious reasons. The second inequality (2 >= 1) is an inclusive OR statement; it's saying that at least one of the two possibilities hold: 2 > 1 or 2 = 1. As 2 > 1 holds, this inclusive OR statement also holds, so 2 >= 1.

OP Tue 17 Aug 2021 16:47:58 No.13529279 Report

Quoted By:

>>13529266
>>13529269
Thanks

OP Tue 17 Aug 2021 17:06:18 No.13529341 Report

Quoted By: >>13529346 >>13529386 >>13529509

>>13529269
Would this still apply in the definition of a limit:
|g(x)-L|=<?
If |x-x_0|<?
Lim(g(x))=L

OP Tue 17 Aug 2021 17:07:19 No.13529346 Report

Quoted By:

>>13529341
Because this would mean that |g(x)-L|>0

Anonymous

Anonymous Tue 17 Aug 2021 17:16:18 No.13529386 Report

Quoted By:

>>13529341
Would what still apply? It need not be true that |g(x) - L| > 0 for any x because g can be the constant function g(x) = L.

Anonymous

Anonymous Tue 17 Aug 2021 17:52:38 No.13529509 Report

Quoted By:

>>13529341
there's a difference between a definition needing <, and a statement saying a number is larger than another one.
In this case both definitions of a limit are equivalent because you can use it on epsilon/2. Then ...<=epsilon/2<epsilon for some possibly smaller delta.

Anonymous

Anonymous Tue 17 Aug 2021 18:08:56 No.13529561 Report

Quoted By:

Holy fuck fuck this board

Anonymous

Anonymous Tue 17 Aug 2021 21:17:43 No.13530317 Report

Quoted By:

>>13529219
In the proof it doesn't matter because if you are given 0<a<b and 0<x<=y then ax<by.
The proof would work just the same with < instead of <=
I admit it is unnecessary and confusing what was done in the proof by using >>13529202 to go from < to <= just to have it turned back into < after multiplication.

Anonymous

Anonymous Tue 17 Aug 2021 21:21:06 No.13530331 Report

Quoted By:

>>13529013
from the first one

Anonymous

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Anonymous Wed 18 Aug 2021 05:53:23 No.13531664 Report

Quoted By:

>>13529013
read this if you actually want to learn calculus.

Anonymous

Anonymous Wed 18 Aug 2021 07:24:31 No.13531775 Report

Quoted By:

>>13529222
drop the class

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