# Problem with the sum of a Fourier series

• Amaelle
In summary, the Fourier sum in the region ##x=2+\epsilon\ ## bounces between ##\log 3 ## and ##\log 2 ##.f

#### Amaelle

Homework Statement
Look at the image
Relevant Equations
Fourier serie
Good day I really don't understand how they got this result? for me the sum of the Fourier serie of of f is equal to f(2)=log(3)
any help would be highly appreciated!

Hi,
Sounds reasonable. However:
what is the sum at ##\ x=2+\epsilon\ ## with ##\ \ 0<\epsilon <<< 1\ ## ?

• Amaelle
as I saw it, I can't plug it the function f(x), so I really have no clue ... The function is periodic, so ##f(2+\epsilon) = f(-{1\over2}+\epsilon)##.
What is ## f(-{1\over2})## ?

• Amaelle
f(-1/2)=f(2)=log(3) as they are periodic, right?

And it jumps down to what when you go a little further ?

so maybe I got you
In this region the value of f(-1/2)=0 so the we are in discountinous stepwise function, in which tthe value of f(2+e)
bounces between 2 and 0?

i mean jump between log3 and 0

## f(-{1\over2}) = 0 ## ? $$\lim_{\epsilon \to 0}f(-\tfrac12 + \epsilon) = \lim_{\epsilon \to 0} |\log(-\tfrac12 + \epsilon)| = \lim_{\epsilon \to 0}\log(\tfrac{2}{1 - 2\epsilon}) = ?$$

## f(-{1\over2}) = 0 ## ? no I meant f(-1/2 +ε) =log(1+ε) =0 because -1/2 does not belong to the definition domain

$$\lim_{\epsilon \to 0}f(-\tfrac12 + \epsilon) = \lim_{\epsilon \to 0} |\log(-\tfrac12 + \epsilon)| = \lim_{\epsilon \to 0}\log(\tfrac{2}{1 - 2\epsilon}) = ?$$
log(2)

log(2)
but sorry just one question
lim f(-1/2 + epsilon)= lim |log(-1/2+epsilon)|=lim -log(-1/2+ epsilon)=lim log(2/(2epsilon-1)) I think there us a small problem with the sign no?

What is ##\bigl |\log {1\over 2} \bigr |## ?

because -1/2 does not belong to the definition domain
Correct, but does it make any difference for the Fourier series ? Do you calculate something different for ##\ (-{1\over 2}, 2]\ ## than for ##\ [-{1\over 2}, 2)\ ## ?

• Amaelle
Correct, but does it make any difference for the Fourier series ? Do you calculate something different for ##\ (-{1\over 2}, 2]\ ## than for ##\ [-{1\over 2}, 2)\ ## ?
not does not

What is ##\bigl |\log {1\over 2} \bigr |## ?
|log(2)|=log(2)

Yes; I hope you got it right: ##\ \bigl |\log{1\over 2}\bigr | = |-\log 2\bigr | = \log 2 \ .##

So at ##x=2## the Fourier sum jumps from ##\log 3 ## to ##\log 2 ##. What's the average ?

##\ ##

• • hutchphd and Amaelle
log(6)/2, thanks a million!

So somehow the subtle difference between ranges ##\ (-{1\over 2}, 2]\ ## and ##\ [-{1\over 2}, 2)\ ## is 'lost in transformation' ##\ ##

• Amaelle
So somehow the subtle difference between ranges ##\ (-{1\over 2}, 2]\ ## and ##\ [-{1\over 2}, 2)\ ## is 'lost in transformation' ##\ ##
yes , thanks a million, that was a nice explanation!