periodic fn

[civic moon]

How to check whether sin((π/2) [x]) and sin((π/3) [x]) are periodic or not and if yes find fundamental period

[civic moon]

Just showing that the whole fn is non periodic because the innermost func I. E (π/2) [x] or π/3 × x is non periodic won't suffice ig

Like sin πx is periodic

[turbid hill]

I dont think they are periodic

Because the innermost functions pi/2 . Floor(x) and pi/3 floor(x) arent periodic either

Hmmm, sin(pix) is periodic but there the rule 2pi/pi =2 is used

This might be a good time for me to revise period in functions

[turbid hill]

Okay i checked my notes, and infact there is only mention that f(g(h(x))) has the same period of h(x) if h(x) is periodic

But it did not say anything about the case where h(x) was not periodic

So the innermost functions not being periodic might actually not be able to determine whether or not a function can be periodic

However there is still a way

Okay the only way i could determine that sin(pi/2 floo(x)) is periodic was by drawing the graph of

It

And it is periodic with period 4

this is actually weird

But my graph is correct

Desmos agrees

The second one

Should be periodic with period 6

however wolfram

But desmos

[civic moon]

K cool

How do u solve these kinda questions via the contradiction method tho

[turbid hill]

what?

[civic moon]

f(x+t) =f(x)
And the putting values for n, x and showing that there's no possible T

[turbid hill]

What is n?

[errant pendant]

U asume that there is a T that f(x)=f(x+t)

It will be like this

Sin(pi/2[x])=sin(pi/2[x+t])

In order to separate t from the floor function ,t has to be an integer

But sin is periodic only if t=2npi

But 2npi is not an integer

Which means that sin(pi/2[x]) is not periodic

[turbid hill]

wait what

[turbid hill]

explain this @errant pendant

[errant pendant]

sin(x)=sin(x+2npi)

and u have

Sin((pi/2)[x])

[turbid hill]

Do you mean t such that f(x)=f(x+t)?

[errant pendant]

yes

[turbid hill]

Basically youre saying assume f(x) has period t okay

[errant pendant]

yes

and come to a contradiction

u agree that

[x+t]=[x]+t?

where t is an integer

[turbid hill]

You mean directly move t outside the floor function? then yes, that is a property of floor

[errant pendant]

yes

but sin is not periodic on t

because it is on 2npi

[turbid hill]

t=2npi, what is n?

[errant pendant]

n is also an integer

[turbid hill]

Are you talking about the sin function in general?

[errant pendant]

yes

in general

[turbid hill]

This is actually wrong then sin(pix) is periodic with period 2, there is no integer n such that 2npi=2

[errant pendant]

no

i mean that

sin(x)=sin(x+2npi)

n is an integer

[errant pendant]

well, thats what i want to say

sin(pi/2[x]+t)=sin(pi/2[x+t])

[turbid hill]

Let me think what youre trying to do

Till then you can explain how 4 is not the period of this function

Also another question why are we forcibly making t an integer?

Just input 4 and i do not see the problem 🗿 @errant pendant

#got em 😼

[humble oxide]

,rccw

[civic moon]

Thanks
@errant pendant @turbid hill @humble oxide

.close

[somber jetty]

Solved

Post marked as solved by @civic moon.

Use .unsolved if this was a mistake.

[errant pendant]

:0

rlly

so it is periodic