/sci/ - Science & Math

[Anonymous No.8844764]

- f(0)=1
- |f(x) - f(y) - x^2 + y^2| ≤ (cosx-cosy)^2

Prove that g(x)=f(x) - x^2 is a constant function

[Anonymous No.8844769]

>>8844764
I see you got it right this time, brainlet.

[Anonymous No.8844771]

>>8844764

uwot m80

[Anonymous No.8844774]

>>8844769
yeah senpai srry for that

[Anonymous No.8844924]

>>8844764
pretty easy desu. Show that the derivative of f(x) is 2x using some well chosen value for y and limits.

[Anonymous No.8846601]

Kek you guys are not as smart as i though

[Anonymous No.8846602]

>>8846601
if it worked using literaly the first strategy I tried >>8844924

I'd say you're pretty dumb. Not doing your homework senpai.

[Anonymous No.8846608]

>>8846602
Already done it and it's not my homework. Also it's not quite what you said just try it nigga.

[Anonymous No.8846610]

>>8846608
I actually made this and so far only 2 of my friends solved this

[Anonymous No.8846653]

>>8846608
I tried it, I didn't type it until I tried, dumbass.

[Anonymous No.8846724]

>>8846653
Did you actually solve it then

[Anonymous No.8846730]

Rewrite the inequality as
[math] |g(x)-g(y)| \leq (cosx-cosy)^2[/math]
Say for x>y and divide by x-y
[math]\left| \frac{g(x)-g(y)}{x-y} \right| \leq \frac{(cosx-cosy)^2}{x-y}[/math]
Take the limit as [math]y\rightarrow x [/math] or w.e of both sides and then
[math]|g'(x)| \leq \lim_{x\rightarrow y}\frac{(cosx-cosy)^2}{x-y} = 0 [/math]
I am a retard and don't know what f(0) = 1 is for.

Also a limit computation that is weird
[math] \lim_{x\rightarrow y}\frac{(cosx-cosy)^2}{x-y} = \lim_{x\rightarrow y}(x-y)\frac{(cosx-cosy)^2}{(x-y)^2} =\lim_{x\rightarrow y}(x-y)\left(\lim_{x\rightarrow y}\frac{cosx-cosy}{x-y}\right)^2 = \lim_{x\rightarrow y}(x-y)\cdot (cos(x)')^2 = 0 [/math]

[Anonymous No.8846734]

>>8846730
forgot to mention it doesn't matter whether y>=x or x>=y because of symmetry.

[Anonymous No.8846743]

>>8846730
the problem was asking find f(x) i just made it easier by asking prove g(x) is a constant function because you have to do that first in order to find f(x) and then use f(0)=1 to find that g(x)=1 and f(x)=x^2 + 1

[Anonymous No.8846768]

>>8846730
f(0)=1 is for the integration constant.
>>8846724
yes

I just wrote y = x+h, divided everything by |h| and came to the same conclusion as >>8846730

It's literaly the first thing anyone would try, and it works.

[Anonymous No.8846776]

>>8846768
do you think if i ask find f(x) instead of prove g(x) is constant then it will be too hard for a high school student?

[Anonymous No.8846968]

>>8846776
I think so. You might ask something like "find the derivative of f" first if they know a definition for derivatives and know something about limits.