Suppose $f(x)=\dfrac{4^x}{4^x+2}$ and prove that $$f\left(\dfrac{1}{1997}\right)+f\left(\dfrac{2}{1997}\right)+\ldots+f\left(\dfrac{1996}{1997}\right)=998.$$  I tried to do the problem but I could not get started. Substituting the value of $x$ seems to be too lengthy.
52780
If   $$f(x) = \frac{4^x}{4^x+2} $$   then find the value of   $$f\left(\frac{1}{1999}\right) + f\left(\frac{2}{1999}\right) + f\left(\frac{3}{1999}\right) +\cdots+f\left(\frac{1999}{1999}\right).$$    I tried it by changing expression to  $$f(x) =1 - \frac{2}{4^x+2}$$ but I am not able to cancel any term.  Is there is any other trick to do it? Thanks in advance.
10003
For $f:\mathbb{R}^n \to \mathbb{R}^m$, if $f(x + y) = f(x) + f(y)$ for then for rational $c$, how would you show that $f(cx) = cf(x)$ holds?  I tried that for $c = \frac{a}{b}$, $a,b \in \mathbb{Z}$ clearly  $$ f\left(\frac{a}{b}x\right) = f\left(\frac{x}{b}+\dots+\frac{x}{b}\right) = af\left(\frac{x}{b}\right) $$ but I can't seem to finish it, any help?
109975
Let $a&gt;0$,$b&gt;0$ and $c&gt;0$. Prove that:    $$\dfrac{a^2+b^2}{a+b}+\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}\le \dfrac{3\left(a^2+b^2+c^2\right)}{a+b+c}.$$     $$\Leftrightarrow \frac{(a^2+b^2)(a+b+c)}{a+b}+\frac{(b^2+c^2)(a+b+c)}{b+c}+\frac{(c^2+a^2)(a+b+c)}{c+a}\leq 3(a^2+b^2+c^2)$$  $\Leftrightarrow 2(a^2+b^2+c^2)+\frac{c(a^2+b^2)}{a+b}+\frac{a(b^2+c^2)}{b+c}+\frac{b(a^2+c^2)}{a+c}\leq 3(a^2+b^2+c^2)$  $\Leftrightarrow \frac{c(a^2+b^2)}{a+b}+\frac{a(b^2+c^2)}{b+c}+\frac{b(a^2+c^2)}{a+c}\leq a^2+b^2+c^2$  $\Leftrightarrow \frac{c(a+b)^2-2abc}{a+b}+\frac{a(b+c)^2-2abc}{b+c}+\frac{b(a+c)^2-2abc}{a+c}\leq a^2+b^2+c^2$  $\Leftrightarrow 2(ab+bc+ac)\leq a^2+b^2+c^2+2abc\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)$  I can't continue. Help me
91147
Find this sum closed form $$f(n)=\sum_{i=1}^{n}\dfrac{\binom{n}{i}}{i}$$  My idea: since $$\dfrac{1}{i}=\int_{0}^{1}x^{i-1}dx$$ so $$f(n)=\sum_{i=1}^{n}\left(\int_{0}^{1}x^{i-1}dx\binom{n}{i}\right)=\int_{0}^{1}\sum_{i=1}^{n}\binom{n}{i}x^{i-1}dx$$  then I can't,Thank you
81465
Show that   $$(p-1)!\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{p-1}\right)\equiv 0\pmod{p^2}.$$   Maybe use this $$\dfrac{1}{k}+\dfrac{1}{p-k}=\dfrac{p}{k(p-k)}$$ and then I can't. Can you help me to prove it?   Thank you.
43823
In general, $\frac{d}{dx}(f(x) \cdot g(x)) \neq  \frac{d}{dx}f(x) \cdot \frac{d}{dx}g(x)$   When does this result hold true? My first try is to use product rule on left side and compare the two sides, but this hasn't helped at all. Any suggestions?
10141
Prove: $\binom{n}{0}F_0+\binom{n}{1}F_1+\binom{n}{2}F_2+\cdots+\binom{n}{n}F_n=F_{2n}$;  I was stuck with this question for a while... Help me please!!! Thanks!!!
101040
I have this functional relation - $$f\left( x \cdot f(y)\right)=x^2 \cdot y^a$$ which I am trying to solve.   I put $x=1$, then I put $f(y)=\dfrac{1}{x}$. I also tried out $y=f^{-1}(1)$, but it doesn't seem to work out. Please help me out. Thank you.
101125
I have tried a little bit to solve the problem which goes as follows:   My intuition says that there exist no $f:\Bbb R\to\Bbb R$ such that $$f(g(x))=x^{2018}\text{ and }g(f(x))=x^{2019}.$$  Note that $$f(g(f(x)))=f(x)^{2018}\implies f(x^{2019})=f(x)^{2018}$$  Similarly, $$g(x^{2018})=g(x)^{2019}$$ Putting $x=1$ in $f(x^{2019})=f(x)^{2018}$, we get $f(1)=f(1)^{2018}$ and thus $f(1)=0$ or $f(1)=1$.  Similarly, putting $x=1$ in $g(x^{2018})=g(x)^{2019}$ we get $g(1)=g(1)^{2019}$ and thus $g(1)=0,1,-1$.  Now, I can't proceed further. Can anybody solve it? Thanks for assistance in advance.
34650
When is $$f=\left(\frac{\sqrt{p+4}}{1-p}-1\right)x^5-3x+\ln5$$ decreasing $\forall\; x$?    Diffrentiating: $$f'=5\left(\frac{\sqrt{p+4}}{1-p}-1\right)x^4-3$$ If $f$ is decreasing, $f'&lt;0$: $$5\left(\frac{\sqrt{p+4}}{1-p}-1\right)x^4-3&lt;0\implies x^4&lt;\frac3{5\left(\frac{\sqrt{p+4}}{1-p}-1\right)}$$ Since if $\left(\frac{\sqrt{p+4}}{1-p}-1\right)\to0^+\implies x^4&lt;+\infty$, which holds forallx, So:$$p=\frac{3\pm\sqrt{21}}2$$which doesn't ply with textbook-answer, since it mentions an domain for p.
30887
I want to prove that: $\ln(x+1)&lt; x$.  My idea is to define:  $f(x) = \ln(x+1) - x$, so:  $f'(x) = \dfrac1{1+x} - 1  = \dfrac{-x}{1+x} &lt; 0, \text{ for }x &gt;0$.  Which leads to $f(x)&lt;f(0)$, so $\ln (x+1)-x&lt;0$.  Is that a valid proof? Any other ideas?  Thanks.
4795
Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ such that $f\left( f(x)^2+f(y) \right)=xf(x)+y$ for all real numbers $x$ and $y$.   Clearly $f(x)=x$ is a solution, check by substitution.  I'm at a loss as to how to show this is the only one or find others.  Any help would be very much appreciated.
38634
$$ f: \Re \longrightarrow \Re \ \in C^{\infty} \\ \exists \ L&gt;0: \ \forall x \in \Re, \forall n\in N \\ |f^{(n)} (x)| \le L \\ f(\frac{1}{n})=0 \ \forall n\in N \\ f(x) \equiv 0  $$  Good morning,  Can you give me an help to make this proof?  This is my thought:   $\lim_{n \rightarrow +\infty } \dfrac{1}{n}=0=\theta$ $f(\theta)=0$ $f^{(n)} (\theta) =0\\f(x)=\sum_{k=0}^n \dfrac{f^{(k)}(x-\theta)^k}{k!}$  $f$ is analytic, the rest of Taylor's expansion is 0, because the derivatives are bounded, $f$ is null.   I have thought this reasoning, but I have difficulties to demonstrate part 3. Can you help me?  Is there any mistake?  Can I use an easier way to make the proof?  Thanks.
99220
Find $f$ if $$ f(x)+f\left(\frac{1}{1-x}\right)=x $$ I think, that I have to find x that $f(x) = f\left(\frac{1}{1-x}\right)$ I've tried to put x which make $x = \frac{1}{1 - x}$, but this equation has no roots in real numbers.
121877
This function seemed to be pretty much straight forward, but my solution is incorrect. I have two questions: 1. Where did I make a mistake? 2. I learned that there are shortcuts for finding a series (substitution / multiplication / division / differentiation / integration of both sides). Is there something that I can apply here?  $$f(x)=\frac{1}{(1-x)^2}  \qquad  f(0)=1$$ $$f'(x)=\frac{2}{(1-x)^3}  \qquad  f'(0)=2$$  $$f''(x)=\frac{6}{(1-x)^4}  \qquad  f''(0)=6$$  $$f'''(x)=\frac{24}{(1-x)^5}  \qquad  f'''(0)=24$$  $$f''''(x)=\frac{120}{(1-x)^6}  \qquad  f''''(0)=120$$ The series is then $$1+\frac{2x}{2!}+\frac{6x^2}{3!}+\frac{24x^3}{4!}+\frac{120x^4}{5!}$$ and when simplified is  $$1+x+x^2+x^3+x^4$$ But the correct answer is  $$1+2x+3x^2+4x^3+5x^4 $$
13721
$$\dfrac{1}{2}\ln(x+7)-(2 \ln x+3 \ln y)$$  Our professor let's us solve this but i do not understand how $\ln$ works. He says it has same properties with $\log$ but i still don't get it. What's the difference of the two?
116151
Define $f(z) = z \ ^ 2 + \lambda$ and $V = \left\{x \in \Bbb C : |x| \gt \frac{1}{2} + \sqrt{\frac{5}{4} + |\lambda|} \right\}$, for some $\lambda \in \Bbb C$.   Then, if $x \in V$ then $|f(x)| \gt |x| +1$.   I tried the following :   Let $x\in V$ then $|f(x)| = |x \  ^ 2 +\lambda | \ge |x |  \ ^ 2 - | \lambda| \gt (\frac{1}{2} + \sqrt{\frac{5}{4} + |\lambda|}) \  ^ 2 - |\lambda| = 1.5 + \sqrt{\frac{5}{4} + |\lambda|} $.  But this doesn't get me anywhere.