Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$f(x) = \log x$ for any real number $x > 0$ and

$$g(n)=\begin{cases} n& \text{if $n$ is even}\\ \tfrac{1}{n}& \text{if $n$ is odd}.\end{cases}$$

for any natural number $n$.

If $x$ is a natural number greater than $1$, then what is the value of

$$f(f(x^{g(10)})) – f(f(x^{g(9)})) + f(f(x^{g(8)})) –\cdots – f(f(x^{g(1)}))$$

What is the best way to this?

share|improve this question
1  
This is not about the value of an equation at all. It's about the value of an expression. I don't know why using these words in standard ways is such a challenge to so many people. –  Michael Hardy Nov 27 '11 at 16:13
    
my mistake.thanks for correcting. –  sunny Nov 27 '11 at 16:32
add comment

2 Answers

up vote 1 down vote accepted
y(n) = f(f(x^g(n)))=f(g(n) * log(x))=log(g(n))+log(log(x))

y(n) = log(n) + log(log(x)) if n is even
y(n) = log(1/n) + log(log(x)) = -log(n) + log(log(x)) if n is odd

now,
y(10) - y(9) = log(10) + log(log(x)) + log(9) - log(log(x)) = log(10) +log (9)
similarly,
y(8) - y(7) = log(8) + log(7)

finaly result y = log(10) + log(9) + ... + log(1) = log(10*9*...*1) = log(10!)
share|improve this answer
add comment

$$f(f(x^{g(n)}))=f(\log x^{g(n)})=f(g(n)\cdot \log x)=\log(g(n))+\log(\log x)$$

So ,

$f(\log x^{g(10)})=\log 10 +\log(\log x)$

$f(\log x^{g(9)})=\log {\frac{1}{9}} +\log(\log x)$ ...etc.

Therefore your expression is equivalent to the :

$\log 10 -\log {\frac{1}{9}}+\cdots+\log 2-\log 1$

share|improve this answer
    
@Swapan,I can't see mistake...which line is incorrect ? –  pedja Nov 27 '11 at 9:20
    
Extremely sorry...I read the question with all + signs :( My bad, your answer is completely fine. So, I removed my wrong comment. –  Tapu Nov 27 '11 at 13:01
    
@Swapan,It's ok... –  pedja Nov 27 '11 at 13:15
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.