/sci/ - Science & Math » Thread #13682359

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Anonymous Sat 25 Sep 17:53:30 2021 No.13682359 View Reply Original Report

Quoted By: >>13682383 >>13682421 >>13682457 >>13682770 >>13683023 >>13685649

training for a master's and this came up
anyone know how to solve it?
It's probably really embarrassing, but I genuinely don't even know how to start

Anonymous

Anonymous Sat 25 Sep 2021 18:02:42 No.13682383 Report

Quoted By: >>13682385 >>13682395

>>13682359
C

Anonymous

Anonymous Sat 25 Sep 2021 18:04:22 No.13682385 Report

Quoted By: >>13682395

>>13682383
why

Anonymous

Anonymous Sat 25 Sep 2021 18:07:04 No.13682395 Report

Quoted By:

>>13682385
I'm not >>13682383, but the square root tends to x^2, so it becomes x, you add it to 3x

Anonymous

Anonymous Sat 25 Sep 2021 18:14:37 No.13682421 Report

Quoted By:

>>13682359
idk move 1 to the other side of the = and start from there

Anonymous

Anonymous Sat 25 Sep 2021 18:19:50 No.13682439 Report

Quoted By:

Factorise by x and use the development of sqrt(1+u) as u goes to 0 (u=1/x here).

Anonymous

Anonymous Sat 25 Sep 2021 18:26:15 No.13682457 Report

Quoted By: >>13682470

>>13682359
$ For \ x \rightarrow +\infty :\\ f(x)=\sqrt{x^2+x}+3x-1 = \sqrt{x^2(1+\frac{x}{x^2})}+3x-1 \sim \sqrt{x^2} +3x-1 = x+3x-1 = 4x-1 = 4x(1-\frac{1}{4x}) \sim 4x +o(1) $

Anonymous

Anonymous Sat 25 Sep 2021 18:30:08 No.13682470 Report

Quoted By: >>13682484

>>13682457
yeah, that seems right
thanks anon

Anonymous

Anonymous Sat 25 Sep 2021 18:33:44 No.13682484 Report

Quoted By:

>>13682470
It's wrong !

Anonymous

Anonymous Sat 25 Sep 2021 18:42:12 No.13682511 Report

Quoted By: >>13682537 >>13682544

For \ x \rightarrow +\infty :\\ f(x)=\sqrt{x^2+x}+3x-1 = x(\sqrt{(1+\frac{1}{x})}+3-\frac{1}{x}) \sim x(1+\frac{1}{2x}+ O(\frac{1}{x^2})+3-\frac{1}{x}) = 4x-\frac{1}{2}+O(\frac{1}{x})

Anonymous

Anonymous Sat 25 Sep 2021 18:47:46 No.13682537 Report

Quoted By:

>>13682511
$\ x \rightarrow +\infty :\\ f(x)=\sqrt{x^2+x}+3x-1 = x(\sqrt{(1+\frac{1}{x})}+3-\frac{1}{x}) \sim x(1+\frac{1}{2x}+ O(\frac{1}{x^2})+3-\frac{1}{x}) = 4x-\frac{1}{2}+O(\frac{1}{x})[\eqn]$

Anonymous

Anonymous Sat 25 Sep 2021 18:49:52 No.13682544 Report

Quoted By: >>13682768 >>13684386 >>13685738

>>13682511
$\ x \rightarrow +\infty :\\ f(x)=\sqrt{x^2+x}+3x-1 = x(\sqrt{(1+\frac{1}{x})}+3-\frac{1}{x}) \sim x(1+\frac{1}{2x}+ O(\frac{1}{x^2})+3-\frac{1}{x}) = 4x-\frac{1}{2}+O(\frac{1}{x}) $

Anonymous

Anonymous Sat 25 Sep 2021 19:59:41 No.13682768 Report

Quoted By:

>>13682544
Thanks Terrance

Anonymous

Anonymous Sat 25 Sep 2021 20:01:40 No.13682770 Report

Quoted By:

>>13682359
uh the masters is in education, right tyrone? Not something serious like civil engineering?

Anonymous

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Anonymous Sat 25 Sep 2021 20:51:39 No.13682929 Report

Quoted By:

Its C

>Why?

I'm a fucking Schizo and never focused in on society when I was younger and fucked my early days. But my gut feeling is never wrong Anon. Hopefully some other cunt can explain it.

Anonymous

Anonymous Sat 25 Sep 2021 21:05:17 No.13682988 Report

Quoted By:

As x approaches infinity the -1 and +x are dwarfed by the x^2 and 3x respectively. So just imagine the formula without those. Then its sqrt of x^2 + 3x which is 4x.

Anonymous

Anonymous Sat 25 Sep 2021 21:05:21 No.13682989 Report

Quoted By:

$\sqrt{x^2-x} - x = (\sqrt{x^2-x} - x ) \frac{\sqrt{x^2-x} + x}{\sqrt{x^2-x} + x} = \frac{x^2 - x - x^2}{ \sqrt{x^2-x} + x} = -\frac{x}{\sqrt{x^2-x} + x} = -\frac{1}{1 + \sqrt{1 - 1/x }} \to -1/2$
Therefore we can replace $\sqrt{x^2 - x}$ with $x + 1/2$ and we get
$x + 1/2 + 3x - 1 + o(1) = 4x - 1/2 + o(x)$

Anonymous

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Anonymous Sat 25 Sep 2021 21:09:45 No.13683011 Report

Quoted By: >>13684329 >>13684335

Here is a proof without using Taylor series or calculus.
$\sqrt{x^2+x} - x = (\sqrt{x^2+x} - x ) \frac{\sqrt{x^2+x} + x}{\sqrt{x^2+x} + x} = \frac{x^2 + x - x^2}{ \sqrt{x^2+x} + x} = \frac{x}{\sqrt{x^2+x} + x} = \frac{1}{1 + \sqrt{1 + 1/x }} \to 1/2$
Therefore
$\sqrt{x^2 - x} = x + 1/2 + o(1)$ . Substituting it back in we find
$\sqrt{x^2 + x} + 3x - 1 = x + 1/2 + o(1) + 3x - 1 = 4x - 1/2 + o(11)$ and we're done.

Anonymous

Anonymous Sat 25 Sep 2021 21:12:47 No.13683023 Report

Quoted By: >>13683080

>>13682359
Factor out the $x^2$ from the radical, then expand: $\sqrt{x^2 + x} = x \sqrt{1 + \frac{1}{x} } \approx x \left(1+ \frac{1}{2x} - \frac{1}{8x^2}\right )$ So $ f(x) \approx x + \frac{1}{2} - \frac{1}{8x} + 3x - 1 = 4x - \frac{1}{2} - \frac{1}{8x}$ Which means $f(x) \sim 4x + o(1) ~ \text{as} ~ x \to \infty$

Anonymous

Anonymous Sat 25 Sep 2021 21:23:22 No.13683066 Report

Quoted By: >>13683080

>masters
>struggling with grade 10 math

Anonymous

Anonymous Sat 25 Sep 2021 21:27:22 No.13683080 Report

Quoted By: >>13683163

>>13683023
Wrong.
>>13683066
:)

Anonymous

Anonymous Sat 25 Sep 2021 21:50:56 No.13683163 Report

Quoted By:

>>13683080
Wrong.

Anonymous

Anonymous Sat 25 Sep 2021 22:22:44 No.13683240 Report

Quoted By: >>13683254

Let's think about this for a second. The question is what does sqrt(x^2+x) approximate when x gets really big?

x = 1: sqrt(x^2+x) = 1.41
x = 10: sqrt(x^2+x) = 10.49
x = 100: sqrt(x^2+x) = 100.49
x = 1000: sqrt(x^2+x) = 1000.49
x = 10,000: sqrt(x^2+x) = 10,000.49

As you can see, it stays VERY close to x so the problem can be approximated by x + 3x-1 = 4x-1

Anonymous

Anonymous Sat 25 Sep 2021 22:28:33 No.13683254 Report

Quoted By:

>>13683240
>it stays VERY close to x
Looks closer to x+1/2 then x.

Anonymous

Anonymous Sat 25 Sep 2021 22:42:38 No.13683296 Report

Quoted By: >>13684329 >>13684331 >>13684335 >>13684419

Sorry for the stupid question, but what's $o(x)$ here? I've never seen that function before.

Anonymous

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Anonymous Sun 26 Sep 2021 07:42:48 No.13684329 Report

Quoted By: >>13684339

>>13683296
>>13683011

It's called a big-o (or small-o, rather) notation.
It is a function on the order of what's in the
argument. So, o(1) is a function containing 1
(or perhaps a mess of terms with 1 in it) and
is in the order of 1 (a constant).

While I'm here, here's my take on the solution
<--- on the pic here

Anonymous

Anonymous Sun 26 Sep 2021 07:44:31 No.13684331 Report

Quoted By: >>13686391

>>13683296
x multiplied by a function that tends to 0 as x goes to infinity.

Anonymous

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Anonymous Sun 26 Sep 2021 07:47:30 No.13684335 Report

Quoted By: >>13684348 >>13684349 >>13686391 >>13686607 >>13687135 >>13687141

>>13683296
>>13683011
It's called a big-o (or small-o, rather) notation.
It is a function on the order of what's in the
argument. So, o(1) is a function containing 1
(or perhaps a mess of terms with 1 in it) and
is in the order of 1 (a constant).

While I'm here, here's my take on the solution
on the pic.

Anonymous

Anonymous Sun 26 Sep 2021 07:49:39 No.13684339 Report

Quoted By:

>>13684329
Are you all retarded ? o(1) is a function that goes to 0, -1/2 isn't an o(1), it's a O(1) though.

Anonymous

Anonymous Sun 26 Sep 2021 07:52:50 No.13684348 Report

Quoted By: >>13684349 >>13684365

>>13684335
Wrong.

Anonymous

Anonymous Sun 26 Sep 2021 07:54:10 No.13684349 Report

Quoted By: >>13684360 >>13684372

>>13684348
>>13684335
Now, why is this incorrect?

Anonymous

Anonymous Sun 26 Sep 2021 07:58:59 No.13684360 Report

Quoted By: >>13684365 >>13684372 >>13687135

>>13684349
Because 1/2 isn't a o(1). An o(1) is a function that goes to 0 asymptotically (as x goes to infinity here).

Anonymous

Anonymous Sun 26 Sep 2021 08:00:35 No.13684365 Report

Quoted By:

>>13684348
>>13684360
That's a fine difference there...

So it should be 4x+O(1), right?

Anonymous

Anonymous Sun 26 Sep 2021 08:03:13 No.13684372 Report

Quoted By: >>13684386 >>13684419

>>13684349
>>13684360
That's a fine difference there...

So, it should be 4x+O(1) right?

Anonymous

Anonymous Sun 26 Sep 2021 08:10:08 No.13684386 Report

Quoted By: >>13684419 >>13684857

>>13684372
Yes, but an O(1) isn't a o(1). A O(1/x) is a o(1) though, so you must develop it to one further order.

Here : sqrt(x^2+x) = x*sqrt(1+1/x) = x*(1+1/(2*x) + O(1/x^2)) = x + 1/2 + o(1)

As it was done in >>13682544

Anonymous

Anonymous Sun 26 Sep 2021 08:33:04 No.13684419 Report

Quoted By: >>13684857 >>13685738 >>13687135

>>13684372
>>13683296
>>13684386
I see...the Taylor series was done to get down to
x^-1 followed by some remainder. Then the
effective remainder O(1/x) is o(1) which is the
more significant remainder here than O(1).

I had to double check, so I agree with this.
Thanks! It's a while since I had numerical
analysis, though...

Anonymous

Anonymous Sun 26 Sep 2021 12:37:42 No.13684857 Report

Quoted By: >>13685020 >>13685738

>>13684386
>>13684419
Dumbo here
How do you evaluate a Taylor series at infinity?

Anonymous

Anonymous Sun 26 Sep 2021 14:01:08 No.13685020 Report

Quoted By:

>>13684857
carefully

Anonymous

Anonymous Sun 26 Sep 2021 17:16:57 No.13685611 Report

Quoted By: >>13685626

C
>rule out (a) and (b) since there are three answers with 4x, and two with 3x, making the former more likely.
>rule out (e) since it lacks the constant every other answer has.
>rule out (d) since it's the only constant out of the four equations that is divided.
And they say these tests are hard.

Anonymous

Anonymous Sun 26 Sep 2021 17:22:45 No.13685626 Report

Quoted By:

>>13685611
Lol wrong. Can't believe /sci/ is getting so filtered by such a simple problem

Anonymous

Anonymous Sun 26 Sep 2021 17:31:33 No.13685649 Report

Quoted By:

>>13682359
>Multiple choice question
Is that their official training? If so maybe consider another university.

Anonymous

Anonymous Sun 26 Sep 2021 17:57:39 No.13685738 Report

Quoted By:

>>13684419
>>13684857
When you evaluate the Taylor series as x
approaches infinity, like the work done on
>>13682544 , the O(x^-1) term goes to zero and
the 4x term obviously becomes infinite. But
instead of letting the series go straight to infinity
like this we stop at a very large but finite number
and state the general behavior of the function.
That is, for large numbers it behaves like
4x - (1/2) with a remainder O(x^-1) that is very
small.

Anonymous

Anonymous Sun 26 Sep 2021 21:38:19 No.13686391 Report

Quoted By: >>13687141

>>13684335
I can't read your vomit
>>13684331
so just a technical way of saying the the -1 is irrelevant compared to x as x -> infty?

Anonymous

Anonymous Sun 26 Sep 2021 22:17:37 No.13686501 Report

Quoted By: >>13686602

The answer is both C D and E are all correct, as the little o doesn't change under addition or subtraction of a constant. Though E is the most simplified form, the others would work just as well.

Anonymous

Anonymous Sun 26 Sep 2021 22:47:36 No.13686602 Report

Quoted By:

>>13686501
>as the little o doesn't change under addition or subtraction of a constant
Yes it does.
$f \in o(g)$ means $f/g \to 0$ .
Try to think a little bit before posting.

Anonymous

Anonymous Sun 26 Sep 2021 22:49:21 No.13686607 Report

Quoted By: >>13687135

>>13684335
Big-o is not the same thing as small-0!
f = O(g) means |f/g| is bounded by a constant in the limit. f= o(g) means f/g goes to zero in the limit. Little-o is a stronger condition. You would know this if you actually looked it up.

Anonymous

Anonymous Mon 27 Sep 2021 02:03:55 No.13687135 Report

Quoted By:

>>13686607
>>13684335
I thought the o's were similar until
I had to look it up after that because of
>>13684360 and then corrected for it by >>13684419.

Anonymous

Anonymous Mon 27 Sep 2021 02:08:11 No.13687141 Report

Quoted By:

>>13684335
>>13686391
Be nice. It's not vomit, it's fancified shit.

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