/sci/ - Science & Math

[Anonymous No.1350636]

Can /sci/ tell me the function for lim x->∞ f(x)=y, and if a=2b, y-f(a)=2(y-f(b))
Thanks

[Anonymous No.1350654]

Done in Latex..
<span class="math">\lim_{x \to \infty} f(x)=1[/spoiler]
a=2b and <span class="math">y-f(a)=2[y-f(b)][/spoiler]

[Anonymous No.1350665]

>>1350654
for example,
<div class="math">f(x)\equiv 1</div>
what is <span class="math">y[/spoiler]?

[Anonymous No.1350666]

wtf am i reading.jpg

[Anonymous No.1350675]

This question is nonsense.

[Anonymous No.1350683]

>>1350654
typo
>>1350636
I don't think there is enough info to determine f(x)

[Anonymous No.1350690]

well the solution won't be unique, for example the constant function <span class="math"> f(x) = y [/spoiler] is trivially a solution

[Anonymous No.1350697]

>>1350690
no
<span class="math"> y \ne 2y [/spoiler]

[Anonymous No.1350702]

>>1350697
try again, retard
y-f(a)=2(y-f(b))
f(a)=y
f(b)=y
Substitute:
y-y=2(y-y)
0=0

This is usually true.

[Anonymous No.1350712]

OP here again
<span class="math">\lim_{x \to \infty} f(x)=y[/spoiler]
a=2b and <span class="math">y−f(a)=2[y−f(b)][/spoiler]

Sorry about the typo, that 1 should have been y.

[Anonymous No.1350715]

>>1350712

f(x) = y for all x is a solution

[Anonymous No.1350726]

>>1350712
what is a and b

[Anonymous No.1350727]

>>1350715
But then it won't fulfil the second set of conditions (unless y=0)

[Anonymous No.1350729]

>>1350726
Arbitrary numbers to be put into the function, a is double b.

[Anonymous No.1350736]

I'll try to translate this to human language.

Find all functions <span class="math">f: f: \mathbb R \rightarrow \mathbb R[/spoiler] such that
1) <div class="math">\lim_{x \to \infty} f(x)=y</div>
2) for any <span class="math">a \in \mathbb R[/spoiler]:
<div class="math"> y - f(2a) = 2(y - f(a))</div>

Is this correct, OP?

[Anonymous No.1350741]

Just don't care about a and b.
f(x) is always y, no matter what value x has. So it can go against infinity or you can insert a or b or whatever, it doesn't matter. It always stays y.
So this anon is right: >>1350702
I think this is more a trick question.

[Anonymous No.1350747]

>>1350727

Yes it will,

LHS: y - f(a) = y - y = 0 as f(x) = y for all x
RHS: 2(y-f(b)) = 2(y-y) = 0 as f(x) = y for all x

I herped so hard I derped

[Anonymous No.1350750]

What about this OP?
f(x)=y-z/x (if z=y, when x=1 f(x)=0, higher z=lower f(x) at x=1)

[Anonymous No.1350760]

>>1350736
If this statement is correct the answer is obvious.
<div class="math">y - f(2a) = 2(y - f(a))</div>
<div class="math">f(2a) = 2 f(a) - y</div>
<div class="math">f(2^n a) = 2^n f(a) - (2^n - 1)y</div>
obviously <span class="math">f(0) = y[/spoiler]
and if for some a f(a) = y+c then <span class="math">f(2^n a) = y + 2^nc[/spoiler]

so f(x) = y for all x

[Anonymous No.1350773]

Sorry, I'm this guy: >>1350741
Is this the task >>1350736? Then I got your question wrong.

But it is still simple then and I still think it's a trick question.
It is fulfilled for f(x) = y.

[Anonymous No.1350865]

Let's hope my LaTeX-fu serves me well:

<div class="math">
Let\: \alpha=\frac{2\pi i}{ln(2)}
\newline
Let\:f(x)=x\cdot (c_1 x^\alpha+c_2 x^{-\alpha})+c_3
\newline
\newline
f(x)\: spirals \:around \:c_3 \:but \:|f(x)|\: grows\:with\:x.
\newline
Assume\:that\:lim_{x\to\infty} f(x)=c_3.
\newline
f(x)\:satisfies\: y-f(2x)=2(y-f(x)).
\newline
Where\:y=c_3.
\newline
c_1=0=c_2\:is\:the\:trivial\:case\:where\:f(x)=c_3=y.
</div>

[Anonymous No.1350867]

>>1350865

NUODAI!?!?

[Anonymous No.1350873]

>>1350865
<div class="math">Let\: \alpha=\frac{2\pi i}{ln(2)}</div>
<div class="math">Let\:f(x)=x\cdot (c_1 x^\alpha+c_2 x^{-\alpha})+c_3</div>
<div class="math">f(x)\: spirals \:around \:c_3 \:but \:|f(x)|\: grows\:with\:x.</div>
<div class="math">Assume\:that\:lim_{x\to\infty} f(x)=c_3.</div>
<div class="math">f(x)\:satisfies\: y-f(2x)=2(y-f(x)).</div>
<div class="math">Where\:y=c_3.</div>
<div class="math">c_1=0=c_2\:is\:the\:trivial\:case\:where\:f(x)=c_3=y.</div>

[Anonymous No.1350879]

>>1350873
btw... ln(2) refers to the principle branch of the log function... c1,c2,c3 are complex numbers

[Anonymous No.1350883]

>>1350873

why are you assuming that limit?

[Anonymous No.1350893]

>>1350883
Because f(x) actually diverges since it grows with x... The average value of f(x) happens to be c3. It's similar to assuming that <div class="math">lim_{x\to\infty} sin(x)=0
</div>
even though this is blatantly wrong. It's convenient to use the average value because it yields the correct answer in the trivial case of c1=c2=0.

[Anonymous No.1350902]

>>1350893

That is some horrible, horrible math right there.

[Anonymous No.1350904]

>>1350893
I should mention that if c1!=c2, the spiral is actually a spiral around a cone-shaped spiral around c3. Quite a beautiful result, really. And I'm only looking at real values of x.

[Anonymous No.1350907]

>>1350902
I know. But I've made my horrid assumptions clear. Take 'em or leave 'em. If you don't like it, then I was only able to find the trivial case.

[Anonymous No.1350910]

>>1350907

You cannot assume something that isn't true! Doesn't matter how nice your results are.

[Anonymous No.1350917]

>>1350910
I can, I'm a physicist.

[Anonymous No.1350921]

>>1350917
read: idiot

[Anonymous No.1350935]

>>1350921
>You cannot assume something that isn't true! Doesn't matter how nice your results are.
http://mathworld.wolfram.com/DeltaFunction.html

[Anonymous No.1350942]

>>1350935

Random link that doesn't disprove the statement.
Nice.

[Anonymous No.1350959]

>>1350942
> zero everywhere except at zero
> infinity at zero
> integrating over zero yields 1... inf*0=1?
> has derivatives
gtfo

[Anonymous No.1350962]

>>1350959

>he still thinks the delta distribution is a function

[Anonymous No.1350970]

>>1350959
Enjoy your heuristic definition

[Anonymous No.1351023]

>>1350962
>>1350970
>are retarded

My point is that we treat it like a function when it's just a useful fantasy (like your mother.)
In particular, I want to point out that the limit definition of a derivative fails to yield a useful result but we still assume that the derivative of the delta function is meaningful and we can derive its properties from the assumption. It's convenient to do so.

[Anonymous No.1351046]

>>1351023

But that's all defined rigorously, look up test functions, so we can make those assumptions. They are not wrong.

What you done with your assumption was not rigorous, it was just wrong.

[Anonymous No.1351060]

>>1351046
> still missing the point
I'm done here.

[Anonymous No.1351065]

>>1351060

Good, this board could do without your bullshit math

[Anonymous No.1351096]

>>1350910
"You cannot assume something that isn't true! "
... yes you can, and you can get useful results from incorrect assumptions, someone mentioned physics jokingly, but thats accurate
... also, see: mathematical axioms

i think that guy was just saying that if the limit was the case, the solution works... he said himself that the limit diverges and you can only get the trivial solution from his answer

[Anonymous No.1351153]

>>1351096

Axioms are self evident and consistent, not wrong.

[Anonymous No.1351185]

>>1350959
>does not know what generalized functions are
>or thinks that <span class="math"> \delta[/spoiler] is singular

[Josef @home !!cvJKiVKmWz2 No.1351234]

Delta "function", tricking highschoolers since 1930

[Anonymous No.1351243]

>>1351234
>>1351185

Finally some back up.

[Josef @home !!cvJKiVKmWz2 No.1351262]

>>1351243
I feel you. Two days ago I had to fight a war against a crowd of people that said the Lorentz force was valid with a wedge instead of a cross.

[Anonymous No.1351494]

>>1351060
>>1351023
New bystander here in the mix. Pretty sure I side with the other anon(s); the delta distribution, not actually a function (hoo-boy it's a "generalized function"), has useful properties when treated like a function ONLY because we established, rigorously, a formal basis for doing so, with heavy analysis on n*g(nx) under the integral, and a bunch of other shit for the "derivatives." In pure mathematics you aren't allowed to indulge in these conveniences without first verifying their validity.

Polite sage, even if it is a true /sci/worthy thread.