Slopes of tangents at roots

Claim: If a polynomial of degree nn has real roots x1,x2,,xnx_1, x_2 , \ldots , x_n such that x1<x2<<xnx_1 < x_2 < \cdots < x_n. If the slopes of tangents at xix_i are mim_i, then 1m1+1m2++1mn=0. \dfrac1{m_1} + \dfrac1{m_2 } + \cdots + \dfrac1{m_n} = 0 .

Proof:

Let P(x)=anxn+an1xn1+an2xn2+an3xn3++a0\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 x1,x2,x3,,xnx_1, x_2, x_3,\cdots, x_n.

Then we can also write it as - P(x)=an(xx1)(xx2)(xx3)(xx4)(xxn)\displaystyle P(x) = a_n (x-x_1) (x-x_2) (x-x_3) (x - x_4) \cdots (x - x_n)

Let a=ana = a_n,

P(x)=a(xx2)(xx3)(xxn)+a(xx1)(xx3)(xxn)++a(xx1)(xx2)(xxn1)\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,

P(x1)=m1=a(x1x2)(x1x3)(x1xn)\displaystyle P'(x_1) = m_1 = a (x_1 - x_2)(x_1 - x_3) \cdots (x_1 - x_n)

P(x2)=m2=a(x2x1)(x2x3)(x2xn)\displaystyle P'(x_2) = m_2 = a (x_2 - x_1)(x_2 - x_3) \cdots (x_2 - x_n)

\vdots

P(xn)=mn=a(xnx1)(xnx2)(xnxn1)\displaystyle P'(x_n) = m_n = a (x_n - x_1)(x_n - x_2) \cdots (x_n - x_{n-1})

1m1=1a(x1x2)(x1x3)(x1xn)\displaystyle \frac{1}{m_1} = \frac{1}{a (x_1 - x_2)(x_1 - x_3) \cdots (x_1 - x_n)}

1m2=1a(x2x1)(x2x3)(x2xn)\displaystyle \frac{1}{m_2} = \frac{1}{a (x_2 - x_1)(x_2 - x_3) \cdots (x_2 - x_n)}

\vdots

1mn=1a(xnx1)(xnx2)(xnxn1)\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 1m1\frac{1}{m_1} and try to expand it using partial fractions.

1m1=A2a(x1x2)+A3a(x1x3)++Ana(x1xn)\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 AiA_is using Partial fractions limit method.

A2=1(x2x3)(x2x4)(x2xn)A_2 = \frac{1}{(x_2 - x_3)(x_2 - x_4)\cdots (x_2 - x_n)} and so on.

Our expression for 1m1\frac{1}{m_1} therefore becomes -

1m1=1(x2x3)(x2x4)(x2xn)a(x1x2)+1(x3x2)(x3x4)(x3xn)a(x1x3)++1(xnx2)(xnx3)(xnxn1)a(x1xn)\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

1m1=1m2+1m3+1m4++1mn\displaystyle \frac{1}{m_1} = \frac{-1}{m_2} + \frac{-1}{m_3} + \frac{-1}{m_4} + \cdots + \frac{-1}{m_n}

QED\large \text{QED}

Note by Kartik Sharma
2 years, 8 months ago

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1 vote

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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 1mi \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 x1=x2 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 - 2 years, 8 months ago

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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 - 2 years, 8 months ago

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Tangent is still the same for complex numbers, namely P(xi) P'(x_i) .

Ah, I see that repeated roots could be an issue given that P(xi)=0 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 xixj 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 - 2 years, 8 months 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 - 2 years, 8 months ago

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@Calvin Lin @Michael Mendrin @Pi Han Goh Can you please check its validity ?

Kartik Sharma - 2 years, 8 months ago

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Yes. It's correct.

For a shorter proof: Consider ddx(ln(P(x)) \dfrac d{dx} ( \ln(P(x)) . Think about the generalized version of the chain rule followed by finding the coefficients of AnA_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 - 2 years, 8 months 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 - 2 years, 8 months ago

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