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Slopes of tangents at roots

Claim: If a polynomial of degree \(n\) has real roots \(x_1, x_2 , \ldots , x_n \) such that \(x_1 < x_2 < \cdots < x_n\). If the slopes of tangents at \(x_i\) are \(m_i\), then \[ \dfrac1{m_1} + \dfrac1{m_2 } + \cdots + \dfrac1{m_n} = 0 . \]

Proof:

Let \(\displaystyle P(x) = a_n x^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + a_{n-3}x^{n-3} + \cdots + a_0\) have roots \(x_1, x_2, x_3,\cdots, x_n\).

Then we can also write it as - \(\displaystyle P(x) = a_n (x-x_1) (x-x_2) (x-x_3) (x - x_4) \cdots (x - x_n)\)

Let \(a = a_n\),

\(\displaystyle P'(x) = a (x - x_2)(x - x_3) \cdots (x - x_n) + a (x - x_1)(x - x_3) \cdots (x - x_n) + \cdots + a (x - x_1)(x - x_2) \cdots (x - x_{n-1})\)

Thus,

\(\displaystyle P'(x_1) = m_1 = a (x_1 - x_2)(x_1 - x_3) \cdots (x_1 - x_n)\)

\(\displaystyle P'(x_2) = m_2 = a (x_2 - x_1)(x_2 - x_3) \cdots (x_2 - x_n)\)

\(\vdots\)

\(\displaystyle P'(x_n) = m_n = a (x_n - x_1)(x_n - x_2) \cdots (x_n - x_{n-1})\)

\(\displaystyle \frac{1}{m_1} = \frac{1}{a (x_1 - x_2)(x_1 - x_3) \cdots (x_1 - x_n)}\)

\(\displaystyle \frac{1}{m_2} = \frac{1}{a (x_2 - x_1)(x_2 - x_3) \cdots (x_2 - x_n)}\)

\(\vdots\)

\(\displaystyle \frac{1}{m_n} = \frac{1}{a (x_n - x_1)(x_n - x_2) \cdots (x_n - x_{n-1})}\)

Let's take \(\frac{1}{m_1}\) and try to expand it using partial fractions.

\(\displaystyle \frac{1}{m_1} = \frac{A_2}{a (x_1 - x_2)} + \frac{A_3}{a (x_1 - x_3)} + \cdots + \frac{A_n}{a (x_1 - x_n)}\)

We can find \(A_i\)s using Partial fractions limit method.

\(A_2 = \frac{1}{(x_2 - x_3)(x_2 - x_4)\cdots (x_2 - x_n)}\) and so on.

Our expression for \(\frac{1}{m_1}\) therefore becomes -

\(\displaystyle \frac{1}{m_1} = \frac{\frac{1}{(x_2 - x_3)(x_2 - x_4)\ldots (x_2 - x_n)}}{a (x_1 - x_2)} + \frac{\frac{1}{(x_3 - x_2)(x_3 - x_4)\ldots (x_3 - x_n)}}{a (x_1 - x_3)} + \ldots + \frac{\frac{1}{(x_n - x_2)(x_n - x_3)\ldots (x_n - x_{n-1})}}{a (x_1 - x_n)}\)

Observe carefully and one will get

\[\displaystyle \frac{1}{m_1} = \frac{-1}{m_2} + \frac{-1}{m_3} + \frac{-1}{m_4} + \cdots + \frac{-1}{m_n}\]

\[\large \text{QED}\]

Note by Kartik Sharma
6 months, 3 weeks ago

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What is the goal that you're trying to achieve? It seems like you want to either 1) Prove the claim or 2) Explain how you arrived at this claim. The styles of writing for these 2 are slightly different. For example, as a proof, the statement of "experience tells us that \( \frac{1}{m_i} \) is a good way to go" seems really weird, since that is exactly what the claim needs. If you are clear on that, you can clean up the writing and make it much easier for your audience to understand what you're saying.

Concerns that I have:

  1. The "substitute \( x_1 = x_2 \)" needs to be justified. They are distinct constants as you stated in the assumption, so I don't understand what you mean by "substitute this number for this distinct number". What are you actually doing here?

  2. In the event that you want to motivate arriving at the claim, you should check the necessity of the conditions. I believe that the assumption of distinct roots and real roots are not needed, as long as we count with multiplicity.

Calvin Lin Staff · 6 months, 3 weeks ago

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@Calvin Lin I have fixed some of it. One concern I have is of double roots. And real roots are required otherwise slopes of tangents do not make sense(at least to me). Kartik Sharma · 6 months, 3 weeks ago

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@Kartik Sharma Tangent is still the same for complex numbers, namely \( P'(x_i) \).

Ah, I see that repeated roots could be an issue given that \( P'(x_i) = 0 \), so the inverse of that doesn't make sense. That would be a good thing for you to work through in the proof, and see if we do need \( x_i \neq x_j \), or what the equivalent statement could be fore repeated roots.

There is very little difference in using real numbers vs complex numbers (BUT not to be confused with complex analysis). Calvin Lin Staff · 6 months, 3 weeks ago

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Is it already discovered? I discovered it while solving a problem on Brilliant.

Maybe there is a geometric proof as well by looking at the final result. Kartik Sharma · 6 months, 3 weeks ago

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@Kartik Sharma @Calvin Lin @Michael Mendrin @Pi Han Goh Can you please check its validity ? Kartik Sharma · 6 months, 3 weeks ago

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@Kartik Sharma Yes. It's correct.

For a shorter proof: Consider \( \dfrac d{dx} ( \ln(P(x)) \). Think about the generalized version of the chain rule followed by finding the coefficients of \(A_n\) via limit method of partial fraction.

I think this is a very nice setup. Maybe you can construct up a problem that utilizes this fact. Don't you think? Pi Han Goh · 6 months, 3 weeks ago

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@Pi Han Goh Yes, that's very true. I should have thought of easier methods of proof before posting it.

Yeah, I am working on it. Nothing much spectacular has come up yet. But I will surely make one. Thanks! Anyways, if you get any good idea for a problem, do share it ! Kartik Sharma · 6 months, 3 weeks ago

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