trigonometric sum

[inland eagle]

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[sly flax]

That is false sin^3(pi/2) is 1 but Im(e^3pi/2) is -1

What you are basically saying is sin^3(x)=sin(3x)

Which is false

You need to first linearize your sin

sin^3(x)=((e^(ix)-e^(-ix))/2i)^3

Yes

Sadly

You need to linearize then you can use the imaginary or real part

[junior ruin]

that's a great exercise in euler's formula, but i'm providing a less complex and more real way to do this for fun

recall a discrete version of integration by parts

,align
\int f \dd g &= fg - \int g \dd f \
\sum_{k=m}^n f_k \Delta g_k &= f_n g_{n+1} - f_m g_m - \sum_{k=m+1}^n g_k \Delta f_{k-1}

[waxen ocean]

vin100

[junior ruin]

here $\Delta$ means ``forward difference'', i.e. $\Delta f_n = f_{n+1} - f_n$.

[waxen ocean]

vin100

[junior ruin]

if u wanna impress others by your math skill level, you can call this "summation by parts"

the basic idea is very simple

to sum $\sum_{k=1}^n a_n$, I write it as $\sum_{k=1}^n (b_{k+1}-b_k)$, so that the sum becomes a telescopic sum which equals to $b_{n+1} - b_1$.

[waxen ocean]

vin100

[junior ruin]

nvm if u dun feel comfortable with this just forget abt this for the moment, and stick with $\sum_{k=1}^n a_n$ as a telescoping sum first

[waxen ocean]

vin100

[junior ruin]

the trick is observe that in two successive terms, the argument (i.e. what's inside the parenthesis (...)) has a common ratio of 1/3.

recall a common technique when dealing with sin² and cos² in (indefinite) integration: make the power disappear. do sth similar for sin³ in this question

i.e. apply the triple angle formula for sines]

[waxen ocean]

vin100

[junior ruin]

to explain the summation by parts formula i'll use a graphical approach

here's a simple diagram for a term on the LHS $$f_k \Delta g_k = {\color{blue!40}f_kg_{k+1}} - {\color{red!40}f_kg_k}.$$

\begin{tikzcd}
& {g_{k+1}} \
{f_k} & {g_k}
\arrow["{+}", color={rgb,255:red,92;green,92;blue,214}, from=2-1, to=1-2]
\arrow["{-}", color={rgb,255:red,214;green,92;blue,92}, from=2-2, to=2-1]
\end{tikzcd}

[waxen ocean]

vin100

vin100

[junior ruin]

to sum up the term from $k = m$ to $k = n$ is to stack up this diagram $n-m+1$ times from bottom to top.

[waxen ocean]

vin100

[junior ruin]

\begin{tikzcd}
& {g_{k+1}} \
{f_k} & {g_k} \
{f_{k-1}} & {g_{k-1}} \
\vdots & \vdots \
{f_{m+1}} & {g_{m+1}} \
{f_m} & {g_m}
\arrow["{+}", color={rgb,255:red,92;green,92;blue,214}, from=2-1, to=1-2]
\arrow["{-}", color={rgb,255:red,214;green,92;blue,92}, from=2-2, to=2-1]
\arrow["{+}", color={rgb,255:red,92;green,92;blue,214}, from=3-1, to=2-2]
\arrow["{-}", color={rgb,255:red,214;green,92;blue,92}, from=3-2, to=3-1]
\arrow["{-}", color={rgb,255:red,214;green,92;blue,92}, from=6-2, to=6-1]
\arrow["{-}", color={rgb,255:red,214;green,92;blue,92}, from=5-2, to=5-1]
\arrow["{+}", color={rgb,255:red,92;green,92;blue,214}, from=4-1, to=3-2]
\arrow["{+}", color={rgb,255:red,92;green,92;blue,214}, from=6-1, to=5-2]
\arrow["{+}", color={rgb,255:red,92;green,92;blue,214}, from=5-1, to=4-2]
\end{tikzcd}

[waxen ocean]

vin100

[junior ruin]

we single out the topmost and bottommost arrows

they give $f_n g_{n+1} - f_m g_m$

[waxen ocean]

vin100

[junior ruin]

the rest is $n-m$ instances of '7'-shaped difference