How do solve an sum of numbers to a common power?

Question

How would solve for $a$ in this equation without using an approximation ? is it possible?

where $x>0$ and $0<a<\infty$

$x=\Sigma _{i=1}^{n} i^a$

for example $120=\Sigma _{i=1}^{6} i^a$ what is $a$ in this equation?

Answer 1

The sum you give yields the Harmonic number, $H_n^{(-a)}$. This can be solved, in Mathematica for instance:

Solve[k == HarmonicNumber[n, -a], a][[1]]

When you plug in $k=120$ and $n=6$ as required by the problem example one finds:

$a = 2.17926794327541855469750380268$

As a check:

$\sum\limits_{i=1}^6 i^a = 119.9999$ (close enough!).

Answer 2

There is not a general closed form but we can use integral estimation

$$\sum_{i=1}^{6} i^a\approx \int_{0.5}^{6.5}x^adx=\left[\frac{x^{a+1}}{a+1}\right]_{0.5}^{6.5}=\frac{6.5^{a+1}}{a+1}-\frac{0.5^{a+1}}{a+1}=120 \implies a\approx 2.175$$

where the value for $a$ needs to be evaluated by numerical methods and by a direct check we obtain

$$\sum_{i=1}^{6} i^{2.175}\approx 119.21$$

Answer 3

If you set $a=\log t$, you can write the question as

$$x=\sum_{k=1}^n k^a=\sum_{k=1}^n t^{\log k}$$

where the RHS is a generalized polynomial (with irrational powers). So this is even less solvable than a polynomial and you can't avoid numerical methods.

The function (of $a$) is strictly growing, comes from $0$ in the negatives, equals $n$ for $a=0$, then tends to $n^a$ asymptotically. A reasonable starting value for Newton's iterations is given by $\log_n x$.

Answer 4

If you want solve for $a$ the equation $$x=\sum_{i=1}^{n} i^a=H_n^{(-a)}$$ you will need a numerical method (as already said in answers) and, for that, you would need some initial estimate.

To get such an estimate, you could consider the asymptotics $$H_n^{(-a)}=n^a \left(\frac{n}{a+1}+\frac{1}{2}+\frac{a}{12 n}+O\left(\frac{1}{n^3}\right)\right)+\zeta (-a)$$ and set the problem as $$x \approx \frac{n^{a+1}}{a+1}$$ for which the solution is given in terms of Lambert function. This will be $$a=-1-\frac{1}{\log (n)}\,W_{-1}\left(-\frac{\log (n)}{k}\right)$$ Since the argument of the function is supposed to be small, you could use for its evaluation $$W(t)=L_1-L_2+\frac{L_2}{L_1}+\frac{L_2(L_2-2)}{2L_1^2}+\frac{L_2(6-9L_2+2L_2^2)}{6L_1^3}+\cdots$$ where $L_1=\log(-t)$ and $L_2=\log(-L_1)$ (as given in the linked page).

For your specific example $(n=6,x=120)$, this would give $a\approx 2.34602$ which is not too bad for an estimate.

Now to polish the root, consider that you look for the zero of function $$f(a)=\sum_{i=1}^{n} i^a-x$$ $$f'(a)=\sum_{i=1}^{n}i^a \log(i)$$ and use Newton method which, starting from the guess $a_0$, will update it according to $$a_{n+1}=a_n-\frac{f(a_n)}{f'(a_n)}$$ Applied to the example, this will give as iterates $$\left( \begin{array}{cc} n & a_n \\ 0 & 2.34602 \\ 1 & 2.19989 \\ 2 & 2.17961 \\ 3 & 2.17927 \end{array} \right)$$

If you do not want to use Lambert function, you could consider that $$\sum_{i=1}^{n} i^a <\sum_{i=1}^{n} n^a=n^{a+1}\implies a_0=\frac{\log (x)}{\log (n)}-1$$ This would be simpler but will require more iterations as shown below $$\left( \begin{array}{cc} n & a_n \\ 0 & 1.67195 \\ 1 & 2.45312 \\ 2 & 2.23205 \\ 3 & 2.18146 \\ 4 & 2.17927 \end{array} \right)$$

Edit

Since function $f(a)$ varies very fast with $a$, it could be better to make it more linear considering instead $$g(a)=\log\left(\sum_{i=1}^{n} i^a \right)-\log(x)$$ $$g'(a)=\frac{\sum_{i=1}^{n} i^a \log(i) }{\sum_{i=1}^{n} i^a }$$ Using the simple estimate, for the worked example, the iterates would be $$\left( \begin{array}{cc} n & a_n \\ 0 & 1.67195 \\ 1 & 2.18953 \\ 2 & 2.17927 \end{array} \right)$$ Just for the fun, let us use $(n=20,x=123456789)$; for this case, the iterates would be $$\left( \begin{array}{cc} n & a_n \\ 0 & 5.21931 \\ 1 & 5.80609 \\ 2 & 5.80465 \end{array} \right)$$