Kvant Math Problem 393
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Problem
Find the sum $$\phi(0)+\phi\left(\dfrac{1}{n}\right)+\phi\left(\dfrac{2}{n}\right)+\ldots+\phi(1),$$ if $\phi(x)=\dfrac{4^x}{4^x+2}$.
M. Levin
Exploration
Let
$$S=\phi(0)+\phi!\left(\frac1n\right)+\phi!\left(\frac2n\right)+\cdots+\phi(1),$$
where
$$\phi(x)=\frac{4^x}{4^x+2}.$$
The arguments form the set
$$0,\frac1n,\frac2n,\ldots,\frac{n-1}{n},1.$$
A natural idea is to pair the terms corresponding to $x$ and $1-x$.
Compute
$$\phi(1-x) =\frac{4^{1-x}}{4^{1-x}+2}.$$
Writing $t=4^x$ gives
$$\phi(x)=\frac{t}{t+2}, \qquad \phi(1-x)=\frac{4/t}{4/t+2} =\frac{2}{t+2}.$$
Hence
$$\phi(x)+\phi(1-x) =\frac{t}{t+2}+\frac{2}{t+2} =1.$$
This identity is extremely suggestive because the set of arguments is symmetric with respect to $x\mapsto1-x$.
To test it, take $n=2$:
$$\phi(0)=\frac13,\qquad \phi!\left(\frac12\right)=\frac12,\qquad \phi(1)=\frac23.$$
The sum is
$$\frac13+\frac12+\frac23=\frac32.$$
This equals $\frac{2+1}{2}$.
For $n=3$,
$$\phi(0)+\phi(1)=1, \qquad \phi!\left(\frac13\right)+\phi!\left(\frac23\right)=1,$$
so the sum is $2$, which equals $\frac{3+1}{2}$.
The only possible subtlety is the midpoint $x=\frac12$ when $n$ is even. Since
$$\phi!\left(\frac12\right)=\frac{2}{2+2}=\frac12,$$
the pairing argument still works. The crucial step is proving rigorously that every term is paired exactly once and handling the midpoint when $n$ is even.
Problem Understanding
We must evaluate
$$S=\sum_{k=0}^{n}\phi!\left(\frac{k}{n}\right), \qquad \phi(x)=\frac{4^x}{4^x+2}.$$
This is a Type C problem, since we are asked to determine the value of a quantity.
The core difficulty is recognizing and exploiting a symmetry of the function. The arguments $\frac{k}{n}$ occur in complementary pairs
$$\frac{k}{n},\quad 1-\frac{k}{n}=\frac{n-k}{n},$$
so the problem reduces to finding a relation between $\phi(x)$ and $\phi(1-x)$.
The expected answer is
$$\frac{n+1}{2},$$
because complementary terms appear to add to $1$.
Proof Architecture
First prove the identity
$$\phi(x)+\phi(1-x)=1$$
for every real $x$; this follows by a direct algebraic computation.
Next show that the set of arguments
$$\left{\frac{k}{n}:0\le k\le n\right}$$
is invariant under the transformation $x\mapsto1-x$, since $\frac{k}{n}$ is paired with $\frac{n-k}{n}$.
If $n$ is odd, partition the $n+1$ terms into $\frac{n+1}{2}$ disjoint complementary pairs, each contributing $1$.
If $n$ is even, partition all terms except $\frac12$ into $\frac n2$ complementary pairs, each contributing $1$, and add the middle term $\phi(\frac12)=\frac12$.
The step most likely to fail under scrutiny is the counting of pairs, especially when $n$ is even and the fixed point $\frac12$ occurs.
Solution
Let
$$S=\sum_{k=0}^{n}\phi!\left(\frac{k}{n}\right), \qquad \phi(x)=\frac{4^x}{4^x+2}.$$
We first establish a functional identity.
For any real $x$, put $t=4^x$. Then $t>0$ and
$$\phi(x)=\frac{t}{t+2}.$$
Also,
$$\phi(1-x) =\frac{4^{1-x}}{4^{1-x}+2} =\frac{4/t}{4/t+2} =\frac{2}{t+2}.$$
Therefore
$$\phi(x)+\phi(1-x) =\frac{t}{t+2}+\frac{2}{t+2} =1.$$
Hence
$$\phi(x)+\phi(1-x)=1 \qquad\text{for all }x.$$
Now consider the arguments
$$\frac{k}{n}, \qquad k=0,1,\ldots,n.$$
For each $k$,
$$1-\frac{k}{n}=\frac{n-k}{n},$$
which is another argument occurring in the sum.
If $n$ is odd, the numbers $k$ and $n-k$ are never equal. Thus the $n+1$ terms split into
$$\frac{n+1}{2}$$
disjoint pairs
$$\left(\frac{k}{n},\frac{n-k}{n}\right).$$
Each pair contributes
$$\phi!\left(\frac{k}{n}\right) +\phi!\left(\frac{n-k}{n}\right) =1.$$
Hence
$$S=\frac{n+1}{2}.$$
If $n$ is even, there is one fixed point,
$$\frac{n/2}{n}=\frac12.$$
All remaining terms form
$$\frac n2$$
disjoint complementary pairs, each contributing $1$. Since
$$\phi!\left(\frac12\right) =\frac{4^{1/2}}{4^{1/2}+2} =\frac{2}{4} =\frac12,$$
we obtain
$$S=\frac n2+\frac12=\frac{n+1}{2}.$$
Thus in both cases
$$\boxed{\displaystyle S=\frac{n+1}{2}}.$$
Verification of Key Steps
The identity
$$\phi(x)+\phi(1-x)=1$$
is the central algebraic fact. Recomputing directly,
$$\phi(1-x) =\frac{4^{1-x}}{4^{1-x}+2} =\frac{4}{4+2\cdot4^x} =\frac{2}{4^x+2},$$
and therefore
$$\frac{4^x}{4^x+2}+\frac{2}{4^x+2}=1.$$
No hidden assumption beyond $4^x>0$ is used.
For odd $n$, the equality
$$k=n-k$$
would imply $2k=n$. Since the left side is even and the right side is odd, this is impossible. Hence every term belongs to exactly one complementary pair.
For even $n$, the equation
$$k=n-k$$
gives $k=\frac n2$. This produces the unique unpaired argument $\frac12$. Evaluating it separately gives
$$\phi!\left(\frac12\right)=\frac12,$$
which supplies the missing half-unit in the total.
Alternative Approaches
Introduce
$$a_k=\phi!\left(\frac{k}{n}\right).$$
The identity
$$a_k+a_{n-k}=1$$
follows immediately from the computation
$$\phi(x)+\phi(1-x)=1.$$
Then
$$2S=\sum_{k=0}^{n}(a_k+a_{n-k}) =\sum_{k=0}^{n}1 =n+1,$$
because the index substitution $k\mapsto n-k$ merely reorders the terms. Dividing by $2$ yields
$$S=\frac{n+1}{2}.$$
This approach is shorter because it avoids separating the cases of odd and even $n$, but it hides the pairing structure. The main solution makes the combinatorial symmetry completely explicit.