Exponential equation

[Albert]

$f(x)=\dfrac {4^x}{4^{x+2}}$

find :$f(\dfrac{1}{2007})+f(\dfrac{2}{2007})+f(\dfrac{3}{2007})+----+
f(\dfrac{2006}{2007})$

 

[topsquark]

Re: exponential equation

$f(x)=\dfrac {4^x}{4^{x+2}}$

find :$\dfrac{1}{2007}+\dfrac{2}{2007}+\dfrac{3}{2007}+----+
\dfrac{2006}{2007}$

Okay, I get the whole add the fraction thing, but what is the f(x) for?

-Dan

 

[MarkFL]

Re: exponential equation

\(\displaystyle f(x)=\frac{1}{16}\)

\(\displaystyle \frac{1}{2007}\sum_{k=1}^{2006}k=\frac{2006\cdot2007}{2\cdot2007}=1003\)

 

[Albert]

Re: exponential equation

sorry a typo happened
$f(x)=\dfrac {4^x}{4^{x+2}}$
find :$f(\dfrac{1}{2007})+f(\dfrac{2}{2007})+f(\dfrac{3}{2007})+----+
f(\dfrac{2006}{2007})$
the original post has been edited as mentioned above

 

[MarkFL]

Re: exponential equation

...

f(x) is not equal to $\dfrac {1}{16}$

I beg to differ:

\(\displaystyle f(x)=\frac{4^x}{4^{x+2}}=\frac{4^x}{4^x\cdot4^2}= \frac{1}{4^2}=\frac{1}{16}\)

 

[Albert]

Re: exponential equation

yes, you are right ,and your solution is very smart !

now this time f(x) is different

$f(x)=\dfrac {4^x}{4^x+2}$

find :$f(\dfrac{1}{2007})+f(\dfrac{2}

{2007})+f(\dfrac{3}{2007})+----+

f(\dfrac{2006}{2007})$

 

[mathworker]

Re: exponential equation

Spoiler

for x<1003,
\(\displaystyle f(x)=\frac{1}{1+2^{1-2x}}\)
\(\displaystyle \sum f(x)=\frac{1}{1+2^{1/2007}}+\frac{1}{1+2^{3/20007}}+...........+\frac{1}{1+2^{2005/2007}}\)
for x>1003,
\(\displaystyle f(x)=\frac{2^{2x-1}}{1+2^{2x-1}}\)
\(\displaystyle \sum f(x)=\frac{2^{1/2007}}{1+2^{1/2007}}+\frac{2^{3/20007}}{1+2^{3/20007}}+...........+\frac{2^{2005/2007}}{1+2^{2005/2007}}\)
hence,
answer is 1003

 

[Albert]

Re: exponential equation

It is very interesting ,we have two different functions ,
but their answers are the same ,this is what I want to
point out .
my solutions are different ,and I will post it later .

 

[Albert]

Re: exponential equation

$f(x)=\dfrac {4^x}{4^x+2}$

find :$f(\dfrac{1}{2007})+f(\dfrac{2}

{2007})+f(\dfrac{3}{2007})+----+

f(\dfrac{2006}{2007})$

note we have :
$f(x)+f(1-x)=\dfrac {4^x}{4^x+2}+\dfrac {4^{(1-x)}}{4^{(1-x)}+2}=1$
$\therefore \,\, f(\dfrac{1}{2007})+f(\dfrac{2006}{2007})=1$
$\,\,\,\,\,\,\,f(\dfrac{2}{2007})+f(\dfrac{2005}{2007})=1$
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$\,\,\,\,\,\,f(\dfrac{1003}{2007})+f(\dfrac{1004}{2007})=1$
all together we have 1003 pairs ,so the answer is 1003