Summations involving functions

[Saurabh]

where r_k are the roots of

,

Find S.
--------------
I don't have a decent approach.
I am very bad at summations.
Although, my friend asked me to assume complex analysis and I still couldn't do it.
A simple hint to this would be appreciated.
Peace.

 

[PeroK]

See what you can do with $n = 3, 4$ first.

 

[Saurabh]

See what you can do with $n = 3, 4$ first.

I am stuck again,
I could only conclude that :-
f(x) = (x-1)(x^(n-1) + x^(n-2) + ... + x - 1) + 1
I really do not know what to do next.

 

[fresh_42]

View attachment 221482

where r_k are the roots of View attachment 221485, View attachment 221484Find S.
--------------
I don't have a decent approach.
I am very bad at summations.
Although, my friend asked me to assume complex analysis and I still couldn't do it.
A simple hint to this would be appreciated.
Peace.

Please post these kind of questions in our homework section, use the template which is automatically inserted there and show at least some kind of own effort!

Here you know what $f(r_k)$ is, so you could use this to express $(1-r_k)^2$.

 

[Saurabh]

Please post these kind of questions in our homework section, use the template which is automatically inserted there and show at least some kind of own effort!

Here you know what $f(r_k)$ is, so you could use this to express $(1-r_k)^2$.

Okay I'll take care of the forum thing. :)
I'm not able to express it as $(1-r_k)^2$
Can you help me out by giving a hint from which i can do.
Or can you show me the approach to the problem?

 

[Saurabh]

@haruspex Sir, please help me with this.
I just can't figure out anything to do.

 

[fresh_42]

Okay I'll take care of the forum thing. :)
I'm not able to express it as $(1-r_k)^2$
Can you help me out by giving a hint from which i can do.
Or can you show me the approach to the problem?

You have $f(r_k)=r_k^n-2r_k+2=r_k^n+2\cdot (1-r_k) = 0$ which calls for a substitution. I don't know whether it'll work, but the sum is another then. I would proceed by building a common denominator and see how the general Vieta can help.

 

[tnich]

Here's an idea. Do you know how to express the coefficients of the polynomial in terms of the roots? It's easy to work out if not. Given a polynomial $x^n + a_{n-1} x^{n-1} + . . . + a_1 x + a_0$ with roots $r_1$, $r_2$, . . . , $r_n$,

$x^n + a_{n-1} x^{n-1} + . . . + a_1 x + a_0 = (x-r_1)(x-r_2) . . . (x-r_n)$

If you multiply out the factors on the right and look at the coefficient of $x^k$ on each side of the equation, you get an expression for each coefficient $a_k$ in terms of the roots. Start with n=3 and look what happens when you divide $a_2$ by $a_0$. Then divide $a_1$ by $a_0$. Now you've got something that might be useful.

 

[StoneTemplePython]

A couple other ideas:

let that polynomial be the characteristic polynomial of some matrix $\mathbf A$. If you know how to telescope a von-neuman series, you might be able to get there.

Alternatively, the problem is awfully close to something that comes up in the derivation of Newton's identities. A non-traditional, but instructive derivation is given by Dan Kalman here:

http://dankalman.net/preprints/NewtonsID.pdf

- - - -
where $\mathbf B: = \big(\mathbf I - \mathbf A\big)^2 = \mathbf I - 2 \mathbf A + \mathbf A^2$

your problem is equivalent to finding

$trace\Big(\mathbf B^{-1} \Big)$

 

[haruspex]

It looks useful to factorise f(x)-1. That allows you to write (1-r_k)^-1 as a finite sum of powers of r_k. Turning those into sums of Vieta ratios still looks hard.
See if https://artofproblemsolving.com/wiki/index.php?title=Newton's_Sums helps.

Edit: Nah, there's a much easier way.
One small hint to go on with - logarithmic differentiation. (I believe this is one way to obtain Newton's identities.)

Edit 2: Forgot to mention... this method might require that the roots are all distinct. Maybe that is why we are told n>2.

 

[StoneTemplePython]

It looks useful to factorise f(x)-1. That allows you to write (1-r_k)^-1 as a finite sum of powers of r_k. Turning those into sums of Vieta ratios still looks hard.
See if https://artofproblemsolving.com/wiki/index.php?title=Newton's_Sums helps.

A related idea:

If you work directly with the elementary symmetric functions $e_k$ in the original polynomial -- ignoring the sign function to simplify for this post--

you find $e_k = 0$ for $k = \{1,2,..., n-2\}$

If you then do a substitution $1 - \lambda_r := \lambda_r$, and get a new polynomial $p^*$

and work through it by revisiting those elementary symmetric sums in context of the substituted polynomial:

$e_0^* =1 = \binom{n}{0}$ (of course)
$e_1^* = \binom{n}{1} + \alpha_1 e_1 = \binom{n}{1}$ because any scaled amount of $e_1$ is still zero.
and with $e_2^*$ is... another binomial coefficient, plus a linear combination of lower order original elementary symmetric sums -- each of which evaluate to zero. (Technical nit: lower order really means a not increased index number, and ignores the zeroth index, as the zeroth index can just be ignored, or we could interpret it as being the binomial coefficient -- i.e. the one thing that isn't cancelled.)

There's some very nice overlapping subproblems going on that continue on for $e_3^*$ and so on... though as is common, it can be rather difficult to come up with the right (non-tedious) notation to show this.

Some extra care is needed to weave in the sign function, and when evaluating $e_{n-1}^*$ and $e_{n}^*$ because they each have original elementary symmetric sums that don't get annihilated.

 

[haruspex]

A related idea:

It really is remarkably easy using the logarithmic differentiation method.

 

[StoneTemplePython]

It really is remarkably easy using the logarithmic differentiation method.

That was my gut instinct on this, but then I went down a strange detour. I re-visited this approach -- the calculus really makes this a lot easier.

 

[Ray Vickson]

It looks useful to factorise f(x)-1. That allows you to write (1-r_k)^-1 as a finite sum of powers of r_k. Turning those into sums of Vieta ratios still looks hard.
See if https://artofproblemsolving.com/wiki/index.php?title=Newton's_Sums helps.

Edit: Nah, there's a much easier way.
One small hint to go on with - logarithmic differentiation. (I believe this is one way to obtain Newton's identities.)

Edit 2: Forgot to mention... this method might require that the roots are all distinct. Maybe that is why we are told n>2.

If the OP means $f_n = x^n - 2x + 2$ (as in the typed version) then for $n=2$ the two roots are $x_1 = 1+i$ and $x_2 = 1-i$, so are distinct. The desired sum is $-2.$

If the OP means $f_n = x^n + 2x - 12$ (as in the attachments) then for $n=2$ the two roots are $x_1 = 1+\sqrt{11} \, i$ and $x_2 = 1 - \sqrt{11} \, i$, again distinct. The desired sum is $-2/11.$

 

[haruspex]

If the OP means $f_n = x^n - 2x + 2$ (as in the typed version) then for $n=2$ the two roots are $x_1 = 1+i$ and $x_2 = 1-i$, so are distinct. The desired sum is $-2.$

If the OP means $f_n = x^n + 2x - 12$ (as in the attachments) then for $n=2$ the two roots are $x_1 = 1+\sqrt{11} \, i$ and $x_2 = 1 - \sqrt{11} \, i$, again distinct. The desired sum is $-2/11.$

Yes, I was thinking of roots of f(x)-1. Thanks.

 

[Ray Vickson]

Okay I'll take care of the forum thing. :)
I'm not able to express it as $(1-r_k)^2$
Can you help me out by giving a hint from which i can do.
Or can you show me the approach to the problem?

If the $r_i$ are the roots of $f_n(x)$ the $s_i = 1-r_i$ are roots of $g_n(x) = f_n(1-x)$. If the $s_i$ are roots of $g_n(x)$ the $t_i = 1/s_i$ are roots of $h_n(t) = t^n g_n(1/t)$. Finally, if the $t_i$ are roots of $h_n(t)$ the $u_i = t_i^2$ are roots of $U_n(u) = h_n(\sqrt{u}) \times h_n(-\sqrt{u}).$ (Despite appearances, this is a polynomial in $u$.)

So, now you have the polynomial $U_n(u)$ whose roots are the $1/(1-r_i)^2$.

 

[tnich]

If the $r_i$ are the roots of $f_n(x)$ the $s_i = 1-r_i$ are roots of $g_n(x) = f_n(1-x)$. If the $s_i$ are roots of $g_n(x)$ the $t_i = 1/s_i$ are roots of $h_n(t) = t^n g_n(1/t)$. Finally, if the $t_i$ are roots of $h_n(t)$ the $u_i = t_i^2$ are roots of $U_n(u) = h_n(\sqrt{u}) \times h_n(-\sqrt{u}).$ (Despite appearances, this is a polynomial in $u$.)

So, now you have the polynomial $U_n(u)$ whose roots are the $1/(1-r_i)^2$.

I followed this up to the last step. It seems to me that the term $h_n(-\sqrt{u})$ would introduce additional roots $2-r_i$. Or maybe I am not correctly interpreting what you mean by the symbol $\times$.

 

[Ray Vickson]

I followed this up to the last step. It seems to me that the term $h_n(-\sqrt{u})$ would introduce additional roots $2-r_i$. Or maybe I am not correctly interpreting what you mean by the symbol $\times$.

The symbol $\times$ means multiplication.

The reason the method works is that if we look at a polynomial $F(x)$ with roots $v_i$ and a polynomial $G(x)$ with roots $v_i^2$, one of the factors of $G$ is $(x-v_i^2)$. This can be written as $(\sqrt{x}-v_i)\,(\sqrt{x}+ v_i) = - (\sqrt{x} - v_i)\, (-\sqrt{x} - v_i)$, which are two of the factors in the product $F(\sqrt{x}) \, F(-\sqrt{x})$.

Note that if we set $\sqrt{x} = s$ we have the product $F(s) \, F(-s)$, This is an even function of $s$ and is a polynomial in $s$, so contains only even powers of $s$. That means that we will have only integer powers of $x$ in the function $G(x) = F(\sqrt{x}) \, F(-\sqrt{x})$ and so it is a polynomial as well.

Of course, one could just take the polynomial $h_n(t)$ whose roots are the $t_i = 1/(1-r_i)$, then apply the method in the link in Post #10 to get the sum of the $1/(1-r_1)^2$, thus by-passing entirely the need for such square-root factoring. This would actually be a more straightforward procedure.

 

[tnich]

I followed this up to the last step. It seems to me that the term $h_n(-\sqrt{u})$ would introduce additional roots $2-r_i$. Or maybe I am not correctly interpreting what you mean by the symbol $\times$.

The symbol $\times$ means multiplication.

The reason the method works is that if we look at a polynomial $F(x)$ with roots $v_i$ and a polynomial $G(x)$ with roots $v_i^2$, one of the factors of $G$ is $(x-v_i^2)$. This can be written as $(\sqrt{x}-v_i)\,(\sqrt{x}+ v_i) = - (\sqrt{x} - v_i)\, (-\sqrt{x} - v_i)$, which are two of the factors in the product $F(\sqrt{x}) \, F(-\sqrt{x})$.

Note that if we set $\sqrt{x} = s$ we have the product $F(s) \, F(-s)$, This is an even function of $s$ and is a polynomial in $s$, so contains only even powers of $s$. That means that we will have only integer powers of $x$ in the function $G(x) = F(\sqrt{x}) \, F(-\sqrt{x})$ and so it is a polynomial as well.

Of course, one could just take the polynomial $h_n(t)$ whose roots are the $t_i = C$, then apply the method in the link in Post #10 to get the sum of the $1/(1-r_1)^2$, thus by-passing entirely the need for such square-root factoring. This would actually be a more straightforward procedure.

Thanks. I understand that now. You could simplify even further by using $g_n(x) = f_n(1-x)$ with roots $1-r_k$, then using quotients of the elementary symmetric sums to find $\sum {\frac 1 {(1-r_k)^2}}$.

 

[Ray Vickson]

Thanks. I understand that now. You could simplify even further by using $g_n(x) = f_n(1-x)$ with roots $1-r_k$, then using quotients of the elementary symmetric sums to find $\sum {\frac 1 {(1-r_k)^2}}$.

How would that quotient method work? I don't recall having seen it.

Anyway, the monic polynomial $G_n(t)$ with roots $t_i = 1/(1-r_i)$ is $G_n(t) = (t-1)^n + 2 t^{n-1}.$ We can determine $\sum t_i^2$ quickly and easily from the coefficients of $t^{n-1}$ and $t^{n-2}$ in $G_n$.

 

[tnich]

How would that quotient method work? I don't recall having seen it.

Anyway, the monic polynomial $G_n(t)$ with roots $t_i = 1/(1-r_i)$ is $G_n(t) = (t-1)^n + 2 t^{n-1}.$ We can determine $\sum t_i^2$ quickly and easily from the coefficients of $t^{n-1}$ and $t^{n-2}$ in $G_n$.

If $a_k$ is the coefficient of $x^k$ in your polynomial $g_n(x)$ from post #16 and $s_i$ are the roots of $g_n(x)$, then $\sum_{i=1}^n {\frac 1 {s_i}}=-\frac {a_1} {a_0}$, $\sum_{i=1}^n {\sum_{j=i+1}^n{\frac 1 {s_is_j}}}$ $=\frac {a_2} {a_0}$, . . .
These are easily combined to get $\sum_{i=1}^n {\frac 1 {s_i^2}}=\sum_{i=1}^n {\frac 1 {(1-r_i)^2}}$

 

[Fred Wright]

I have a suggestion. Write $$f(x)=x^n-2x+2=\prod_{s=1}^n (x-r_s) \\ f'(x)=nx^{n-1} -2=\prod_{s=1}^n (x-r_s) \sum_{s=1}^n \frac {1} {(x-r_s)} \\ \sum_{s=1}^n \frac {1} {(x-r_s)} = \frac {nx^{n-1} -2} {x^n-2x+2} \\ f''(x)=n(n-1)x^{n-2}=f'(x) \sum_{s=1}^n \frac {1} {(x-r_s)}- f(x)\sum_{s=1}^n \frac {1} {(x-r_s)^2}$$
I won't go any further but to say that now you can consider f(1) and f'(1) for substitution in the last equation.

Peace
Fred