# Triple sum $\sum\limits_{a=1}^{\infty} \sum\limits_{b=1}^{\infty} \sum\limits_{c=1}^{\infty} \frac{\cos a \cos b \cos c}{a^2 + b^2 + c^2}$

We have poor water heating system in our countryside house (currently it takes 4 hours to warm up the water), and my father has decided to improve it; he bought a water tank and placed it up in the attic, and the last stage is to buy a correct heating element. So he gave me a call and asked to calculate which heating element to buy (more precisely, he asked what electric power does the heating element need to make the descending water hot in 10 minutes; hot was defined as 40 Celsius). The water tank is huge and the heater is placed at one of its dead ends; it wouldn't be correct to say the temperature is uniform and does not vary from one point to another.

The only person for whom I can abandon my work is my father, so I took a standard heat equation: $$\frac{\partial T}{\partial t} = a^2 \left( \frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} + \frac{\partial^2 T}{\partial z^2} \right) + f (x,y,z,t)$$

with initial and boundary conditions $$T(x,y,z,0) = T_0, \hskip 20 pt T(x,y,z,t)\Big|_{\text{boundary}} = T_0$$

and with $$f(x,y,z,t) = \frac{P}{c\rho} \delta(x - x_0) \delta (y - y_0) \delta (z - z_0)$$ for this case (point-like heater located at $$(x_0, y_0, z_0)$$, $$P$$ is heater electric power, $$c$$ is water specific heat capacity, $$\rho$$ is water density), $$a = \sqrt{\kappa/c\rho}$$ is water thermal diffusivity, $$\kappa$$ is water thermal conductivity.

After a tough computation I got the solution:

$$T (x,y,z,t) = T_0 + \frac{P}{lwh \kappa} \sum\limits_{n=1}^{\infty} \sum\limits_{u=1}^{\infty} \sum\limits_{v=1}^{\infty} \frac{\sin \left(\frac{\pi n x}{l} \right) \sin \left(\frac{\pi u y}{w} \right) \sin \left(\frac{\pi v z}{h} \right) \sin \left(\frac{\pi n x_0}{l} \right) \sin \left(\frac{\pi u y_0}{w} \right) \sin \left(\frac{\pi v z_0}{h} \right)}{\left( \frac{\pi n}{l} \right)^2 + \left( \frac{\pi u}{w} \right)^2 + \left( \frac{\pi v}{h} \right)^2 } \cdot$$ $$\cdot \left(1 - \exp\left( - \left( \frac{\pi n a}{l} \right)^2 t - \left( \frac{\pi u a}{w} \right)^2 t - \left( \frac{\pi v a}{h} \right)^2 t \right) \right)$$

So guys, if you will ever install a water tank in your attic, you know where to look. Here $$l$$, $$w$$, $$h$$ are length, width and height of the tank.

My father's question was "which $$P$$ do I need for achieving $$T_{\text{hot}} = 40$$ Celsius from $$T_0 = 20$$ Celsius in $$t = 10$$ minutes in a point $$(x,y,z) = (x_1, y_1, z_1)$$?" To resolve it, to find the $$P$$, I need to numerically compute this triple sum somehow.

Here go my thoughts on the subject:

• the triple sum $$\sum\limits_{n=1}^{\infty} \sum\limits_{u=1}^{\infty} \sum\limits_{v=1}^{\infty} \frac{1}{n^2 + u^2 + v^2}$$ does not converge; you need to have a power of at least $$(3+\varepsilon)$$ in denominator

• trigonometric functions in the numerator really matter: $$\sum\limits_{n=1}^{\infty} \frac{1}{n}$$ does not converge but $$\sum\limits_{n=1}^{\infty} \frac{\cos n}{n}$$ does

• one can utilise the formula for sine product ($$\sin \left(\frac{\pi n x_1}{l}\right) \sin \left(\frac{\pi n x_0}{l}\right) = \dots$$) and rewrite the numerator into 8 terms of three cosines

• thus, what I'm interested in is the sum $$\sum\limits_{n=1}^{\infty} \sum\limits_{u=1}^{\infty} \sum\limits_{v=1}^{\infty} \frac{\cos n \cos u \cos v}{n^2 + u^2 + v^2}$$

• clearly, for numeric computation (e.g. in Mathematica) it would be wonderful to find the precise $$N$$, $$U$$, $$V$$ such that $$\sum\limits_{n=1}^{\infty} \sum\limits_{u=1}^{\infty} \sum\limits_{v=1}^{\infty} (\dots) \approx \sum\limits_{n=1}^{N} \sum\limits_{u=1}^{U} \sum\limits_{v=1}^{V} (\dots)$$

• There's a decent identity $$\int\limits_0^{\infty} \frac{sin at}{a} e^{-bt} dt = \frac{1}{a^2 + b^2}$$, which is proved by integration by parts twice. It allows you to split $$\frac{1}{a^2 + b^2}$$ into factors, into multipliers. Given this split, computing $$\sum\limits_{a=1}^{\infty} \sum\limits_{b=1}^{\infty} \frac{1}{a^2 + b^2}$$ is relatively easy (well, one can calculate both sums and then prove the resulting integral does not converge). However, I failed to split $$\frac{1}{a^2 + b^2 + c^2}$$ in this manner.

• With Fourier transform (see wiki, end of the page, row 502)

$$\frac{1}{a^2 + b^2 + c^2} = \iiint\limits_{\mathbb{R}^3} \frac{e^{-2\pi i (a \xi_a + b\xi_a + c\xi_c)}}{\xi_a^2 + \xi_b^2 + \xi_c^2} d\xi_a d\xi_b d\xi_c$$

the triple sum $$\sum\limits_{a=1}^{\infty} \sum\limits_{b=1}^{\infty} \sum\limits_{c=1}^{\infty} \frac{\cos a \cos b \cos c}{a^2 + b^2 + c^2}$$ could be reduced to ordinary sums such as $$\sum\limits_{a=1}^{\infty} \cos a \cdot e^{-2\pi i a \xi_a}$$. However, the latter sum does not converge.

• Your triple sum certainly converges (albeit slowly), but I doubt we can get a closed form. How accurately would you need it if a numerical approximation were offered? – J.G. Apr 22 at 9:24
• Put yourself on my place: I need to tell my father the correct value for $P$. Then he could be the happiest man on Earth, since I helped him (and go and buy the heater he needs, though this is less important). In case you're interested in water tank parameters, heater location and reference point, they are $l=1$, $w = 0.6$, $h = 5/12$, $(x_0,y_0,z_0) = (0.01, w/2, w/2)$, $(x_1,y_1,z_1) = (l/2,w/2,0.02)$. All values are given in meters. – Alex Garkoosha Apr 22 at 9:31
• I'm not sure I understand the geometry of your setup. You seem to be modelling the heat transfer within the tank itself? Are you warming up all the water in the tank and then using it? For this you won't need the heat equation, it's simply a matter of computing the energy needed to warm all the water in the tank. – Winther Apr 22 at 10:39
• If the tank it full ($m = lwh\rho \sim 250$kg) then you need $P = \frac{c \Delta T m}{t} \approx 35kW$ where $c = 4184$ J/kg/K, $\Delta T = 20$ K and $t = 600$ sec. – Winther Apr 22 at 10:48
• The tank is not that large (50cm - 100cm) and water leads heat fairly efficiently and $600$ sec is a fairly long time so I doubt the answer will be very different. In any case that is a lower bound so just add a bit more to be safe (and account for heat loss etc.). Anyway if you want to do it properly I would simplify the geometry: make it a tube with radius $\sim 50$cm and length $100$cm (choose radius to make volume the same). This is a good approximation and will hopefully lead to simpler expressions. – Winther Apr 22 at 11:32