How many real roots does the following polynomial equation have?

Question

$P_n(x)=1+2x+3x^2+4x^3+....+(n+1)x^n$ Also given: n is an odd number.

Well; from the polynomial equation, it is clearly understood that all the roots have to be negative.

Also, there has to be at least 1 root; that is clear if we put $x=1$.

It would help if you could tell me how to proceed...

Answer 1

Since all coefficients of $P_n(x)$ is non-negative. $P_n(x) \ge P_n(0) = 1 > 0$ for all $x > 0$.
Independent of whether $n$ is even or odd, $P_n(x)$ doesn't have any non-negative root.

For odd $n$, notice

$$(x-1)^2 P_n(x) = (n+1)x^{n+2} - (n+2) x^{n+1} + 1$$

At $x = -r < 0$, we have $$(r+1)^2 P_n(-r) = 1 -((n+1)r + (n+2))r^{n+1}$$ Notice RHS is strictly decreasing as $r$ increase from $0$. Together with the fact RHS is $1$ at $r = 0$ and goes to $-\infty$ as $r \to \infty$. There is one and only one $r > 0$ where $(r+1)^2P_n(-r)$ vanishes.

From this, we can conclude $P_n(x)$ has exactly one real root for odd $n$.

Answer 2

As your function is a partial sum for the binomial series of $(1-x)^{-2}$, consider \begin{align} (x-1)P_n(x)&=-(1+x+x^2+...+x^n)+(n+1)x^{n+1}\\ (x-1)^2P_n(x)&=1-(n+2)x^{n+1}+(n+1)x^{n+2} \end{align}

From the last form you can see by the Descartes rule of signs that there is exactly one negative root for odd $n$.

In a first, rather heuristic, approximation, the roots of the last expression are close to the roots of the two binomials in it, that is, close to $\frac{n+2}{n+1}$ and the roots of $x^{n+1}=\frac1{n+2}$. Remove from this root set the first one and $\sqrt[n+1]{\frac1{n+2}}$ as these correspond to the double root at $1$ of the factor $(x-1)^2$. Then there remains a real root around $x=-1$ with a closer approximation to $-\sqrt[n+1]{\frac1{2n+3}}$ while the other are close to complex unit roots.

More precisely, set $x=qv$ with $q^{n+1}=1$, $q\ne 1$ and $v\approx 1$. Then one approximation of the equation is $$0\approx 1-(n+2-(n+1)q)v^{n+1}$$ which has a solution close to $1$ of approximately $$ v\approx ((n+2)-(n+1)q)^{-1/(n+1)}\approx(2n+3)^{-1/(n+1)}(1+\frac1{2n+3}(q+1)) $$

phi=np.linspace(0,2*np.pi,160); 
for k in range(4,22): 
    n=2*k-1; rts = np.roots(range(n+1,0,-1)); 
    plt.plot(rts.real, rts.imag, '+', label="n=%d"%n); 
    r=(2*n+3)**(-1./(n+1)); q = np.exp(1j*phi); 
    x = r*q*(1+(q+1)/(2*n+3)); 
    plt.plot(x.real, x.imag, lw=0.2)

plt.plot(np.sin(phi),np.cos(phi), lw=0.4); plt.legend(); plt.grid(); plt.show()