I am having this equation: 1/(x−a_1)+1/(x−a_2)+⋯+1(/x−a_n)=0 where a_1<a_2<⋯<an are real numbers. Now I want to prove with the intermediate value theorem that this equation has n−1 solutions in the real numbers.

Ethen Frey 2022-07-22 Answered
I am having this equation:
1 x a 1 + 1 x a 2 + + 1 x a n = 0
where a 1 < a 2 < < a n are real numbers.
Now I want to prove with the intermediate value theorem that this equation has n 1 solutions in the real numbers.
My thoughts:
With a 1 < a 2 < < a n , you can see that every summand gets smaller than the summand before.
My other thought was that about the n-summands, with the intermediate value theorem you know that every zero (point) is in the interval and is located between the n-summands. So there are n 1 solutions for this equation!
Questions:
How can I prove my thoughts in a formal correct way? (Are my thoughts generally correct?)
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Answers (2)

emerhelienapj
Answered 2022-07-23 Author has 14 answers
Let's call your function f. We'll look at it moving left to right along the line. When x < a 1 , f ( x ) < 0, because all the summands are negative. So there are no roots there.
Now let's look at ( a 1 , a 2 ). In particular let's look at x a 1 + . All but the first term is negative, but the first term is going to + while the others are remaining finite. So lim x a 1 + f ( x ) = + . In particular, f is positive somewhere on ( a 1 , a 2 ). Similarly we get that f is negative somewhere on ( a 1 , a 2 ) by considering x a 2 .
You can take it from here.

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Kade Reese
Answered 2022-07-24 Author has 3 answers
We can use the intermediate value theorem with this beginning.
On ( a 1 , a 2 ), approaching a 1 gives us x 1 with f ( x 1 ) > 0. Approaching a 2 gives us x 2 with f ( x 2 ) < 0. Then the intermediate value theorem gives us x 3 ( x 1 , x 2 ) with f ( x 3 ) = 0.

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