Roots of sum of reciprocals

Below I will generalize the problem for the number of solutions to $\sum_{k=1}^{n} \frac{1}{x-k} = 1$ in the interval $[1,n]$ and set $n=5$ .

First, we begin by subtracting 1 from both sides and defining the LHS as a function.

$f(x)=[\sum_{k=1}^{n} \frac{1}{x-k}]-1=0$

It's clear that there are asymptotes at each integer value in the interval $[1,n]$ . Since we are looking for solutions in this interval, we need to find roots to the equation only in between each integer value inside the desired interval. Finding $f'(x)$ yields our desires solutions.

$f'(x)=\sum_{k=1}^{n} -\frac{1}{(x-k)^2}$

It's clear that every term is negative no matter what value you set $x$ to be, meaning that the slope of our function is always negative. Therefore, because we have asymptotes at $k$ and $k+1$ , our function must assume all values from $(-\infty,\infty)$ , and each value of $x$ maps to a unique value of $f(x)$ in the interval $[k,k+1]$ . That means that there must be a solution to $f(x)=0$ and this solution is unique in the interval $[k,k+1]$ . $f(x)$ has $n-1$ such intervals, meaning that $f(x)$ must have $n-1$ roots, and these intervals are all contained within the larger interval $[1,n]$ , so we're done.

We have shown that the equation $\sum_{k=1}^{n} \frac{1}{x-k} = 1$ must have $n-1$ solutions inside the interval $[1,n]$ . Finally, letting $n=5$ yields our final answer: $n-1=5-1=\boxed{4}$

The equation becomes:

x^5 - 20 x^4 + 145 x^3 - 480 x^2 + 724 x - 394 = 0

x1 = 8.390541233048952

x2 = 4.540394425688128

x3 = 3.434736666495783

x4 = 2.356631854844214

x5 = 1.277695819922924

Substituted back to L.H.S. of original equation and found all give the R.H.S. of 1, since 8.390541233048952 is out of interval between 1 to 5, there remained with only 4 real roots as introduced above.

Let $f(x)=(\displaystyle\sum_{n=1}^{5}\frac{1}{x-n})-1$ . Now take a small number $\epsilon>0$ . It is very easy to show that $\displaystyle\lim_{\epsilon \to 0} f(1+\epsilon) \to +\infty$ and $\displaystyle\lim_{\epsilon \to 0} f(2-\epsilon) \to -\infty$ . So there exists at least one real root of f(x) in $[1,2]$ . Now $f'(x)=-\displaystyle\sum_{n=1}^{5}(\frac{1}{x-n})^2$ and it is not zero for any real finite value of x as it is sum of inverse of squares. So there is no maximum or minimum of f(x). This means there is exactly one root of f(x) in $[1,2]$ . By same logic one can show there exists exactly one root in each of the intervals $[2,3],[3,4],[4,5]$ . So there are 4 roots of f(x) in $[1,5]$ .

1. Write as xf = 1 (excuse for writing functions where they belong :) ), then xf - 1 = 0.
2. Differentiate and we have f' < 0, so f always has a negative slope.
3. Next evaluate xf-1 at zero - we get 0f -1 = -1 - 1/2 - 1/3 - 1/4 -1/5 being lazy I let DrRacket do the summation: > (+ -1 -1/2 -1/3 -1/4 -1/5 -1) -197/60
4. So xf-1 starts life with x=0 at a bit less than -3, with poles at 1, 2, 3, 4 and 5.
5. Each time xf-1 dives, it has to come up for air (otherwise how would it be able to dive again??) and we see that xf-1 = -infinity at 1-, 2-, 3- 4- and 5-, and +infinity at1+, 2+, 3+ and 4+ and is continuous monotonic strictly decreasing over the ranges 1+ to 2-, 2+ to 3-, 3+ to 4-, and 4+ to 5-, so there are 4 roots.

Multiply by (x-1)(x-2)(x-3)(x-4)(x-5) both sides of the equation. The roots are still the same. Now you have a polynomial of degree 5 so the first expression has 5 roots. In (-inf,1) the function always decreases and in (5,inf) always increases. With that info it is a trivial check gives that there are no roots on (-inf,1) and there is only one root in (5,inf). It is also a trivial check by Bolzano that there is one root in each of these intervals (1,2), (2,3), (3,4), (4,5). There are no more roots by the Fundamental Theorem of Algebra so 4 is the answer

Before we begin, first we will prove a useful and relevant result.

Given $P(x) = \prod_{i=1}^n (x - x_i)$

Prove that $\frac{P'(x)}{P(x)} = \sum_{i=1}^n \frac{1}{x - x_i}$

$ln(P(x)) = \sum_{i=1}^n {ln(x - x_i)}$

Differentiate both sides wrt. x

$\frac{P'(x)}{P(x)} = \sum_{i=1}^n \frac{1}{x - x_i}$

Moving back to the original problem:

Let $P(x) = (x-1)(x-2)(x-3)(x-4)(x-5)$

$\therefore$ It follows that $\frac{P'(x)}{P(x)} = 1$ is out equivalent problem

$P'(x) = P(x)$

$\therefore$

...It follows, that I have no idea what I'm doing...

Only that I took something completely unrelated and crammed it into the problem.

Anyone have any ideas?

It is very clear that x=1, 2, 3, 4, 5 are asymptotes of the function f(x) $\displaystyle f(x)= \sum_{k=1}^5\dfrac 1 {x-k} -1.\\Investigating~near ~ the~asymptotes,\\f(n+.01)>0, f(n+.99)<0, for~~ n=1, ~2,~3,~4,\\\text {In each of the FOUR interval between 1, 2, 3, 4 .. f(x) changes signs.} \\\text{So there are FOUR real roots for x [1,5]}$ $\color{#D61F06}{\text {Surprisingly, the notes I had posted has vanished!! }}$
In the notes I had given exact value of the fifth root between 8 and 9. I had also reasoned whey the fifth root is not between 5 and 6 as is the case of other roots. When the function is graphed it shows only one intersection with x-axis between asymptotes.
It may be noted that a reciprocal graph of the type $y= \dfrac 1 x \text{always have ONLY ONE intersection between asymptotes. }$

The given function can be modified to : (x - 2) (x - 3) (x - 4)(x - 5) + (x - 1) (x - 3)(x - 4)(x - 5) + (x - 1) (x - 2) (x - 4)(x - 5) +(x - 1) (x - 2) (x - 3)(x - 5) + (x - 1) (x - 2)(x - 3) (x - 4) - (x - 1) (x - 2) (x - 3)(x - 4)(x - 5) = f(x)

When x = 1, f(x) is +ve
x = 2, f(x) is -ve
x = 3, f(x) is +ve
x = 4, f(x) is -ve
x = 5, f(x) is +ve

Since there are 5 sign changes, 4 roots are there.