11. Checking the formulas for Fourier Series with WolframAlpha program

Rotating Fourier Series

Chapter 11. Checking the formulas for Fourier Series with WolframAlpha program

Chapter11.1 Introduction
My main goal is to convince the Reader of the following formulas, which are thoroughly discussed in chapter 7.2. They are not easy, but I hope the large amount of examples with animations helped a bit. It is known that the center of gravity scn of the trajectory for a given rotational speed nω0 is almost the nth harmonic. The formula of Fig. 11-1b is generally intuitive, especially for scn=(0,0) when the harmonic for nω0 does not exist. Yes, it was intuitive, but not calculated. Now we will use the WolframAlfa program from the Internet. You don’t have to install anything or pay anything. By the way, you will get acquainted with this program. It is a pity that as a student in the 60’s I could not use it.

Fig. 11-1
Formulas to remember from chapter 7.
We will check them with the WolframAlfa program. Then they will be more understandable.
Fig. 11-1a
Trajectory F(njω0t) The point moves on the real axis Re Z according to the periodic function f(t) and the plane Z rotates with ω=-nω0. In this way, the trajectory F(njω0t) is drawn for any n.
Fig. 11-1b
Formula for the center of gravity scn of the trajectory F(njω0t). The subscript function is the trajectory F(njω0t). The integral (divided by 2π) can be treated as a point scn, average distant from the trajectory F(njω0t) from the beginning z=(0,0) at the time  T=2π sec.
Fig. 11-1c
Formula for the constant component c0=a0 of the Fourier series. Classic mean of the periodic function f(t) when n=0 in the formula b.
Fig. 11-1d
Formula for the remaining nth coefficients of the Fourier cn series for ω=1ω0, ω=2ω0… ω=nω0. Otherwise, the nth complex amplitude of the harmonic hn(t)
Fig. 11-1e
A complex number scn as a vector with components an and jbn
Fig. 11-1f
scn in the algebraic version scn=an+jbn
Fig. 11-1g.
cn in the exponential version cn=|cn|exp (jϕn) The module |cn| is clearly visible and the phase ϕn for pulsations ω=nω0.
Fig. 11-1h
nth harmonic hn(t) as a sum of cosine and sinus components.
Fig. 11-1i
nth harmonic hn(t) as cosine with phase ϕ. Module |cn| is a “pythagoras” of an and bn, and tg(ϕ)=bn an.

Chapter 11.2 The WolframAlfa Program
We will use it to check the formula for the centers of gravity scn of the known trajectories. Most of these types of programs have one drawback. To solve the problem, you need to define it strictly. You make a mistake in putting the period and already receive an “error” or other difficult to understand message. WolframAlfa is so wise that all he needs to do is ask a general question and he will give many answers. All the more, the more general the question was. You will choose the answer that best suits you. Thanks to this, you do not have to remember all the instructions and you can completely focus on the problem.
For example, we want to solve the equation 2x+3=7.
1. Call up the program from the Internet.
2. Enter 2x+3=7 in the box.
3. The program doesn’t really know what you mean. Just in case, it will draw a graph and give the solution x=-2. You cared about the latter i.e.  x=-2
Note:
After calling the program, you will exit this blog. To return to it, click the Windows return arrow –>.
Click https://www.wolframalpha.com do what the picture tells you to do.
Fig. 11-2
How did WolframAlfa solve the equation 2x+7=3?
For you, the most important solution is solution x=–2. And the chart as an additional answer will not hurt.
It was a primer. I wonder how WolframAlfa will cope with integration? Especially with integration of vectors or complex numbers.
Let us now calculate the centers of gravity scn and by the way the nth harmonics for several known periodic functions f(t). Before that, you had to take the center of gravity on faith. Now you calculate them.

Chapter 11.3 Center of gravity sc1 of the trajectory F(jnω0t)=F(-1j1t)=1exp(-1j1t)
Chapter 11.3.1 Introduction
So for the constant function f(t)=1 with p. 7.4.1 Chapter 7.
It is a periodic function, because the function repeats every period of T. Not only that, what any T period!

Fig. 11-3
Complex function 1exp(-1j1t) as a trajectory and its “center of gravity” scn.
Fig. 11-3a
Complex function 1exp(-1j1t) as a rotating vector.
Click. During the period T=2π/ω≈6.28sec, the vector will perform one turn. What if we added successive vectors during the rotation?
Fig. 7-3b
A circle as a trace of a rotating vector, i.e. a trajectory 1exp(-1j1t).
Fig. 11-3c
The formula for the center of gravity scn of the trajectory of a function when n=1 and f(t)=1.
Even without calculations, it can be seen that the center of gravity is sc1=(0,0). Will the formula Fig. 11-1b confirm this?

Chapter 11.3.2 WolframAlfa
Let us calculate the center of gravity sc1 of the trajectory F(-1j1t)=f (t)exp (-1j1t).
Click on https://www.wolframalpha.com and do what the picture tells you to do.
Instead of laboriously typing an instruction, you can just copy it.
integrate 1*exp(-i1t)/(2pi) from t=0 to t=2pi
and paste into the window.
In other words, you will paste into the window the integral formula Fig. 11-1b written in WolframAlfa language.
Remember that the imaginary number for WolframAlfa is “mathematical i” and not the “electric j” used in the blog.

Fig. 11-4
Center of gravity sc1 = (0,0) trajectory F (-1j1t)=1exp(-1j1t)
As we expected, the center of gravity sc1 = 0 or more precisely sc=(0,0), because it is a complex number. Wolfram is nice! For example, he changes the instruction from the window into a human, or mathematical language. It can be seen by the arrow “3.You will get it. ” Returning to the result sc1=0, what is the result of it?  Well, the amplitude of the first harmonic c1 for the harmonic 1ω0=1/sec is zero, because cn=2scn. So there is no harmonic with pulsation ω=1 / sec, nor for any other pulsation ω. This is what we expected and the only component of the constant function f(t)=1 is the constant component c0=a0=1.

Chapter 11.4 Centers of gravity scn of trajectory F(-njω0t)=0.5cos(4t exp(-njω0t) for n=0,1,4 and ω0=1/sec
Chapter 11.4.1 Introduction
This is an abbreviated version of Chapter 4, in which there were 9 versions of this trajectory for n=0…9. Now there are only 3 versions for n=0,1,4 and only these formulas will be checked in the accounting
Let’s throw the function f(t)=0.5cos(4t) into the centrifuge.
Let us turn the rotation to n=0, n=1 and n=4. So for the speed ω=0 (the centrifuge is standing!), ω=-1/sec and ω=-4/sec.
There will be 3 trajectories:
n=0–>ω = 0–>centrifuge is standing–>F(-0j1t)=0.5cos (4t)
n=1–>ω=-1/sec ->F(-1j1t)=0.5cos(4t)exp(-1j1t)
n=4–>ω=-4/sec ->F(-4j1t)=0.5cos(4t)exp (-4j1t)

Fig. 11-5
Three trajectories F(-nj1t)=0.5cos(4t)exp(nj1t) for n=0, 1 and 4
ω = 0
The centrifuge is standing –>sc0=(0,0)
ω=-1sec
Rotating ω=-1sec –>sc1=(0.0)
ω=-4/sec
Rotating ω=-4/sec –>sc4=(0.25.0)
The above parameters sc0, sc1 and sc4 although intuitive, were taken as “my word of honor”. Now we will calculate them with the WolframAlfa program. The results should be the same.

Chapter 11.4.2 Center of gravity sc0, that is for ω=0, that is the constant component sc0=c0=a0.
So when the centrifuge is stopped, because n=0. Then the trajectory F(-0j1t)=f (t)=0.5cos(4t).
The animation Fig. 11-4a ω = 0 shows that the trajectory is a horizontally swinging line according to of the function f (t)=0.5cos (4t). Its center of gravity is evidently sc0=(0,0). Will WolframAlfa confirm it? Let’s put f(t)=0.5cos (4t) into the formula Fig. 11-1c and calculate this integral.
Click on https://www.wolframalpha.com and do as the picture tells you.
Enter or paste integrate 0.5cos(4t)/(2pi) from t=0 to t=2pi into the box

Fig. 11-6
Calculation of the center of gravity sc0 for F(0j1t)=0.5cos (4t)exp(-0j1t)=0.5cos (4t)
sc0 = 0
Note that although the base period f(t) is T=π/2, the result of sc0=0 for the integration limits T=π/2 and T=2π will be the same!
The limits of integration in the formula Fig. 11-1c can be any and the result will be the same for e.g. T=0.234 sec, T=5.27 sec or T=2π sec as in the above formula! It is only important that T is a period of the function. This remark is more general and applies to the Fourier Series formulas in Fig. 11-1b. Some give arbitrary limits of integration to T, and others, like me, specific T=2π. After all, the Fourier coefficients for example of a square wave do not depend on the period T, but on the shape of this wave!
The center of gravity sc0 for the trajectory of each function f (t) is also a constant component of this function and is the coefficient c0=a0 of the Fourier Series. Unlike the other Fourier coefficients, i.e. c1, c2 … cn, the coefficient c0 is always a real number.

Chapter 11.4.3 Center of gravity sc1, that is for ω=-1/sec.
So sc1 of the trajectory F(-1j1t)=0.5cos(4t)exp (-1j1t) from Fig. 11-5 ω=-1/sec. We are looking for a harmonic with pulsation ω=1/sec from the function f(t)=0.5cos (4t). It is obvious that it does not exist. After all, the only harmonic f (t) is itself, i.e. 0.5cos (4t). It also results from Fig. 11-4 ω=-1 sec, where sc1=(0,0). Will WolframAlfa confirm it?
Click on https://www.wolframalpha.com and do as the picture tells you.
Enter or paste integrate [0.5cos (4t)/(2pi)]exp(-1i1t) from t=0 to t= pi into the window.

Fig. 11-7
Center of gravity sc1=0 trajectory 0.5cos (4t)exp (-1j1t).
Amplitude c1=0 for the harmonic ω=1/sec because c1=2sc1=0. So there is no such harmonic.
As you saw in Chapter 4, the centers of gravity scn for all ω other than 4/sec (that is, from ω of the function 0.5cos (ωt)) are zero. It is like, as the classic used to say, obvious obviousness. After all, the function 0.5cos (4t) has only one harmonic and it is itself! The other harmonics are zero.

Chapter 11.4.4 Center of gravity sc4, that is for ω=-4/sec.
So sc4 trajectory F(-4j1t)=0.5cos (4t)exp (-4j1t) from Fig. 11-4 ω=-4/sec.
We are looking for a harmonic with pulsation ω=4/sec from the function f (t) = 0.5cos (4t). I think everyone sees this harmonic. It is itself 0.5cos(4t).
The function f (t)=0.5cos(4t) rotates clockwise at a speed of ω=-4/sec. Note that f(t) and the centrifuge have ω= 4/sec (albeit opposite signs)! This resulted in a shift of the center of gravity from (0.0) to sc4=(+0.25.0).
To check the accounting click https://www.wolframalpha.com and do what the picture tells you.
Type or paste integrate[0.5cos (4t)]*exp(-i4t)/(2pi) from t=0 to t=2pi into the box.

Fig. 11-8
Center of gravity sc4=0.25 or more precisely sc4=(0.25,0) of the trajectory 0.5cos (4t)exp (-4j1t)
It will turn out that for each spin speed ω of the function 0.5cos (4t)/(2pi) different from ω=4/sec the center of gravity scn=(0,0), and only for ω=4/sec there is a non-zero sc4=(+ 0.25.0)!
Acc. to Fig. 11-1d  c4=2sc4=(+ 0.5,0) that is a4=0.5 and b4=0
Acc. to  Fig. 11-1h  h4(t)=0.5cos (4t).

Chapter 11.5 Centers of gravity scn of trajectory F (-njω0t) = [1.3+0.7cos(2t)+0.5cos (4t)] exp (-njω0t) that is F (-njω0t) for n = 0,2,3,4 and ω0=1/sec
Chapter 11.5.1 Introduction
It is an abbreviated version of Chapter 5, in which there were 9 trajectories for n = 0… 9. Now there will only be 4 for n=0,2,3 and 4.
Let’s put the function f (t)=1.3+0.7cos (2t)+0.5cos (4t) into the centrifuge
Let the rotation be at ω=0, ω=-2 sec, ω=-3/sec, and ω=-4/sec.
4 trajectories will be created:
n=0–>ω=0–> centrifuge is standing–>F(0j1t)=1.3+0.7cos(2t)+0.5cos (4t)
n=1–>ω=-2/sec -> F(-1j1t)=[1.3+0.7cos(2t)+0.5cos(4t)]exp (-2j1t)
n=3–>ω=-3/sec -> F(-3j1t)=[1.3+0.7cos(2t)+0.5cos(4t)]exp(-3j1t)
n=4–>ω=-4/sec -> F(-4j1t)=[1.3+0.7cos(2t)+0.5cos(4t)]exp (-4j1t)

Fig. 11-9
Four trajectories F(-nj1t)=[1.3+0.7cos (2t)+0.5cos(4t)]exp(-nj1t) for n=0, 1, 2 and 4 and their centers of gravity scn.
The waveforms during one period T=2π sec. The waveforms ω=-2/sec and ω=-4/sec are “drawn on top of each other” and therefore seemingly stopped after 1π sec.
ω=0 The centrifuge is standing–> sc0=(1.3,0)
ω=-2/sec                                             –>sc2=(0.35.0)
ω=-3/sec                                             –>sc3=(0,0)
ω=-4/sec                                            –>sc4=(0.5,0)
The centers of gravity of the trajectories scn are fairly intuitive, but were included in Chapter 5 without justification.
Now we will calculate them by of the formula Fig. 11-1b with the WolframAlfa program. The results should be the same.

Chapter 11.5.2 Center of gravity sc0 for ω=0
So sc0 of the trajectory F(-0j1t)=[1.3+0.7cos (2t)+0.5cos (4t)] from Fig. 11-9 ω = 0. The centrifuge is standing. The trajectory is the horizontally swinging line accl to F(-0j1t)=f (t)= 1.3+ 0.7cos (2t)+0.5cos(4t). Its center of gravity is sc0=1.3=(1.3,0).
Let us calculate sc0=c0=a0 that is the constant component acc. to Fig. 1-11c.
Click on https://www.wolframalpha.com and do what the picture tells you to do.
Enter or paste the WolframAlfa instruction for the integral, i.e.
integrate[1.3+0.7*cos (2t)+0.5*cos (4t)]/(2pi) from t = 0 to t = 2pi

Fig. 11-10
Calculation of the center of gravity sc0 for F(0j1t)=[1.3+0.7cos(2t)+ 0.5cos(4t)]/(2pi)
The center of gravity sc0=(1.3,0)=1.3 for the non-rotating trajectory of the function f(t)
is its constant component, i.e. the coefficient c0=a0=1.3 of the Fourier Series.

Chapter 11.5.3 Center of gravity sc2 for ω=2/sec
So sc2 trajectory F(-2j1t)=[1.3+0.7cos(2t)+0.5cos(4t)]*exp(-2j1t) for Fig. 11-9 ω=-2/sec
Click https://www.wolframalpha.com 
Enter or paste integrate [1.3+0.7*cos(2t)+0.5*cos(4t)]*exp (-2i1t)]/(2pi) from t = 0 to t = 2pi 
Fig. 11-11
Calculation of the center of gravity sc2 for F (2j1t)=[1.3+0.7cos (2t) + 0.5 cos (4t)]*exp (-2j1t)/(2pi)
Center of gravity sc2=0.35=(0.35.0) , i.e. the 2nd harmonic amplitude is c2=2(0.35.0]=(0.7,0). i.e. a2=0.7 b2=0
The second harmonic is h2(t)=0.7cos(2t) acc. Fig. 11-1h.

Chapter 11.5.4 Center of gravity sc3 for ω=3/sec
So sc3 trajectory F(-3j1t)=[1.3+0.7cos(2t)+0.5cos(4t)]*exp(-3j1t) for Fig. 11-9 ω=-3/sec
Click https://www.wolframalpha.com 
Enter or paste integrate [1.3+0.7cos(2t)+0.5cos(4t)]*exp(-3j1t)]*exp (-3i1t)]/(2pi) from t=0 to t=2pi Fig. 11-12
Calculation of the center of gravity sc3 of the trajectory [1.3+0.7cos (2t)+0.5cos(4t)]*exp (-3j1t)/(2pi)
Center of gravity sc3=0=(0,0) , i.e. 3rd harmonic amplitude c3=2*0=0.There is no third harmonic. There is also no 5,6,7 … harmonic.
You can see this in the f(t) function itself, of course, but you can check it with WolframAlfa.

Chapter 11.5.5 Center of gravity sc4 for ω=4/sec
So sc4 trajectory F(-4j1t)=[1.3+0.7cos(2t)+0.5cos(4t)]*exp(-4j1t) for Fig. 11-9 ω=-4/sec
Click https://www.wolframalpha.com 
Enter or paste integrate [1.3+0.7*cos(2t)+0.5*cos(4t)]*exp(-4i1t)]/(2pi) from t = 0 to t = 2pi 

Fig. 11-13
Calculation of the center of gravity sc4 for F (4j1t)=[1.3+0.7cos (2t) + 0.5 cos (4t)]*exp (-4j1t)/(2pi)
Center of gravity sc4=0.25=(0.25.0) , i.e. the 4th harmonic amplitude is c4=2(0.25.0]=(0.5,0). i.e. a4=0.5 b4=0
The fourth harmonic is h4(t)=0.5cos(4t) acc. Fig. 11-1h.

Chapter 11.6 Centers of gravity scn of trajectory f(nj1t)=0.5cos(4t-30°)*exp(-4j1t) for n = 0.2,3.4 and ω0=1/sec that is with completely complex center of gravity scn
Until now, the centers of gravity scn of the harmonics were real numbers, eg Fig. 11-9 ω=-2/sec–->sc2=(0.35.0)= 0.35. They are also complex numbers, but with zero imaginary components. What if the cosine function has a phase shift?
So let’s throw into the centrifuge e.g. f(t)=0.5cos(4t-30 °).
Let us turn the rotation to n=0, n=1 and n = 4.
There will be 3 trajectories:
n=0–>ω=0–> centrifuge is standing–> F(-0j1t)=0.5cos (4t-30 °)
n=1–>ω=-1/sec->F(-1j1t)=0.5cos(4t-30 °)*exp (-1j1t)
n=4–>ω=-4/sec ->F(-4j1t) = 0.5cos (4t-30 °)*exp (-4j1t)

Fig. 11-14
F(nj1t)=0.5cos(4t-30°)*exp(-jnt) for n=0,1 i 4
The animation lasts T≈6.28sec.
ω=0
F(0j1t)=0.5cos(4t-30°)
The difference between the animation Fig. 11-5 ω=0 is minimal, but try to notice it.
ω=-1/sec.
F(4j1t)=0.5cos(4t-30°)*exp(-j1t)
The trajectory is rotated by (I suspect?) -30° relative to Fig. 11-5 ω=-1/sec. Center of gravity sc0=(0,0). This means that there is no harmonic of the function f(t)=0.5cos(4t) for ω=1/sec. Also for any other, except for ω=4/sec.
ω=-4/sec.
F(4j1t)=0.5cos(4t-30°)*exp(-4j1t)
The center of gravity sc4 is a full-fledged complex number sc4=(a,b) where:
a=0.25cos(-30°)≈+0.217
b=0.25sin(-30°)≈-0.125
sc4=(a,b)≈(+0.217,-0.125)=+0.217-j0.125
So the complex fourth harmonica h4(t)≈2*sc≈2*(+0.217-j0.125)*exp(j4t). It corresponds to the function f(t)=0.434cos(4t)+0.25sin(4t)=0.5cos (4t-30°) according to the formulas Fig. 11-1h and Fig. 11-1i What does WolframAlfa say?
Click https://www.wolframalpha.com.
Enter or pastej integrate [0.5cos(4t-pi/6)]/(2pi)*exp(-4i1t) from t=0 to t=2pi
Note  -30° =π/6

Fig. 11-15
sc4=0.216506-j0.125
Such center of gravity sc4 of the trajectory F(4j1t)=0.5cos(4t-30°)*exp (-j4t) was calculated by WolframAlfa. A result similar to ours, only more accurate. Note that sc4 is a full complex number. This is the case when the cosine/sinus type function has a non-zero
phase shift ϕ. Here ϕ=30°=-π/6.
Fun fact
Compare the animations Fig. 11-13 ω=-1/sec and ω=-4/sec.
ω=-1/sec
Four Leaf Clover
ω=-4/sec
It is also some kind of a leafy clover except that each leaf is a circle and is drawn “on top of each other”. 

Chapter 11.7 Centers of gravity scn trajectories
F(-njω0t)=[0.5+1.08cos (1t-33.7 °)+0.72cos(3t+33.7°)+0.45cos(5t-26.6 °)]*exp(-njω0t)
for n = 0,1,2…8 and ω0=1/sec
Chapter 11.7 Introduction
The function f(t) with the period T=2πsec looks like this and you don’t see its components 3 cosineoids with phases ϕ.
Otherwise WolframAlfa theoretical doesn’t need to know the formula  f(t). For him, the f(t) diagram alone is enough.

Fig. 11-16
f(t)=0.5+1.08*cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°)
In Chapter 6 we examined 9 trajectories F(-njω0t) of this function f(t) for n=0,1,2,… 8 and ω0=1/sec. It is interesting because due to the phase shifts ϕ, the centers of gravity scn are completely complex numbers. The function f(t) is equivalent to the following, in which each harmonic has been decomposed into a cosine and a sinusoidal component.
f(t)=0.5+0.9cos(1t)+0.6sin(1t)+0.6cos(3t)-0.4sin(3t)+0.4cos(5t)+0.2sin(5t).
e.g. 0.9*cos(1t)+0.6*sin(1t)=1.08*cos(1t-33.7°)
It is the same function, but complex Fourier coefficients, or complex Fourier amplitudes, are easily determined here. From them, harmonics are read as h1(t), h3(t) and h5(t) waveforms with sine/cosine components.
c1=0.9-j0.6—>h1(t)=0.9*cos(1t)+0.6*sin(1t)
c3=0.6+j0.4–>h3(t)=0.6*cos(3t)-0.4*sin(3t)
c5=0.4-j0.2—>h5(t)=0.4*cos(5t)+0.2*sin(5t)
Note:
Both forms of the function f(t) are equivalent, but for the calculations we will take the version with “cosines and sines”
We will put the functions f(t) into the centrifuge with different speeds nω0=-n/sec for n=0,1,2,3 and 5. We will check if the calculated centers of gravity scn are the same as in Chapter 6.

Chapter 11.7.2 Centers of gravity scn of trajectory F(-njω0t)=f (t)exp(-njω0t) for n = 0, 1,2, 3, 5 and ω0=1 sec
Let’s put the function from Fig. 11-15 into the centrifuge. I would like to remind you that it can also be presented in the sine/cosine version, i.e: f(t)=0.5+0.9cos(1t)+0.6sin(1t)+0.6cos(3t)-0.4sin(3t)+0.4cos(5t)+0.2sin(5t).
Let’s turn on the rotation for n=0, 1, 2, 3 and 5. So for the speeds ω=0 (the centrifuge is standing!), ω=-1/sec, ω=-2/sec, ω=-3/sec and ω=-5/sec. The formation of 4 trajectories:
ω=0–>centrifuge is standing–>F(-0j1t)=f(t)
ω=-1/sek                                      –>F(-1j1t)=f(t)exp(-1j1t) 
ω=-2/sek                                      –>F(-2j1t)=f(t)exp(-2j1t)
ω=-3/sek                                      –>F(-3j1t)=f(t)exp(-3j1t)
ω=-5/sek                                      –>F(-5j1t)=f(t)exp(-5j1t)

Fig. 11-17
5 trajectories F(-nj1t)=f t)exp (nj1t) for n = 0,1,2,3 and 5 ω = 0 and their centers of gravity scn.
ω=0
The centrifuge is standing and sc0=(+ 0.5.0)
ω=-1/sec 
Rotating ω=-1/sec –>sc1=(+0.45,-0.3)
ω=-2/sec 
Rotating  ω=-2/sec –>sc2=(0,0)
ω=-3/sec 
Rotating  ω=-3/sec –>sc3=(+0.3,+0.2)
ω=-5/sec
Rotating  ω=-5/sec–>sc5=(+0.2,-0.1)
Let me remind you that the vector scn is almost the nth harmonic of the periodic function f(t). More specifically, the doubled vector i.e. 2scn is the complex amplitude of the nth harmonic. In the period T=2π sec, each trajectory is drawn by a changing vector and scn is the average of these changing vectors and therefore their sum is divided by 2π.

Chapter 11.7.3 Center of gravity sc0, i.e. for ω=0
Trajectory F(0j1t)=f(t) from Fig. 11-16 ω=0.
The centrifuge is standing. The trajectory is the horizontally swinging line in Fig. 11-17 ω=0 acc. to of the function F(0j1t)=f(t). Its center of gravity is sc0=(+0.5,0)=+0.5.
Let’s put f(t) into the formula Fig. 11-1c.
Click https://www.wolframalpha.com and do what the picture tells you to do. Enter or paste the WolframAlfa instruction.
integrate [0.5+0.9*cos(1t)+0.6*sin(1t)+0.6*cos(3t)-0.4*sin(3t)+0.4*cos(5t)+0.2*sin(5t)]/(2pi) dt from t=0 to t=2pi

Fig. 11-18
Calculation of the center of gravity sc0 for F(0j1t)=f(t)/(2pi)
You see sc0 as a surface of the f(t) ((divided  by 2π). Note that they are plus and minus sufaces. The center of gravity sc0=(0.5,0)=a0=0.5 for the non-rotating trajectory of the function f (t) is its constant component, i.e. the coefficient c0=a0 of the Fourier Series.

Chapter 11.7.4 Center of gravity sc1 for ω=1/sec.
Trajectory F(-1j1t)=f(t)exp(-1j1t) from Fig. 11-17 ω=-1/sec
Click https://www.wolframalpha.com. Type or paste into the box
integrate [0.5+0.9*cos(1t)+0.6*sin(1t)+0.6*cos(3t)-0.4*sin(3t)+0.4*cos(5t)+0.2*sin(5t)]*exp(-1i1t)]/(2pi) dt from t=0 to t=2pi

Fig. 11-19
Trajectory  F(-1j1t)=f(t)*exp(-1j1t)
Note:
In the window, you only see the part of the expression you pasted.
The calculated center of gravity is sc1=0.45-j0.3
The complex amplitude of the first harmonic is 2sc1=0.9-j0.6
So the first harmonic according to Fig. 11-1h and Fig. 11-1i is:
h1(t)=0.9cos(1t)+0.6sin(1t)=1.08*cos(1t-33.7°).
Success! WolframAlfa perfectly filtered the first harmonic from the periodic function f(t). We expect the same in the following chapters.

Chapter 11.7.5 Center of gravity sc2 for ω=2/sec.
Trajectory F(-2j1t)=f t)exp(-2j1t) from Fig.11-17 ω=-2/sec
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Type or paste into the box
integrate (0.5+0.9*cos(1t)+0.6*sin(1t)+0.6*cos(3t)-0.4*sin(3t)+0.4*cos(5t)+0.2*sin(5t))*exp(-2i1t)/(2pi) dt from t=0 to t=2pi

Fig. 11-20
Trajectory F(-2j1t)=f(t)*exp(-2j1t)
sc2=0
That is, there is no harmonic with pulsation ω=2/sec 

Chapter 11.7.6 Center of gravity sc3 for ω=3/sec.
Trajectory F(-3j1t)=f t)exp(-3j1t) from Fig. 11-17 ω=-3/sec
Click https://www.wolframalpha.com.
Type or paste into the box
integrate (0.5+0.9*cos(1t)+0.6*sin(1t)+0.6*cos(3t)-0.4*sin(3t)+0.4*cos(5t)+0.2*sin(5t))*exp(-3i1t)/(2pi) dt from t=0 to t=2pi

Fig. 11-21
Trajectory F(-3j1t)=f(t)*exp(-3j1t)
The calculated center of gravity is sc3=0.3+j0.2
The complex amplitude of the third harmonic is 2sc1=0.6-j0.4
So the third harmonic according to Fig. 11-1h and Fig. 11-1i is:
h3(t)=0.6cos(3t)-0.4sin(3t)≈0.72*cos(3t+33.7°)..

Chapter 11.7.7 Center of gravity sc5 for ω=5/sec.
Trajectory F(-5j1t)=f t)exp(-5j1t) from Fig. 11-17 ω=-5/sec
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Type or paste into the box
integrate (0.5+0.9*cos(1t)+0.6*sin(1t)+0.6*cos(3t)-0.4*sin(3t)+0.4*cos(5t)+0.2*sin(5t))*exp(-5i1t)/(2pi) dt from t=0 to t=2pi

Fig. 11-22
Trajectory F(-5j1t)=f(t)*exp(-5j1t)
The calculated center of gravity is sc5=0.2-j0.1
The complex amplitude of the third harmonic is 2sc1=0.4-j0.2
So the fifth harmonic according to Fig. 11-1h and Fig. 11-1i is:
h5(t)=0.4cos(5t)+0.2sin(5t)≈0.45*cos(5t-26.6°)

Chapter 11.7.8 “Center of gravity” scn trajectory F (-njω0t)=f (t)exp(-njω0t) for n= 4,6,7,8 and ω0=1/sec
There are still centers of gravity sc4, sc6, sc7 and sc8 i.e. for n=4,6,7 and 8. They are zero, i.e. there are no harmonics for these pulsations n*ω0. I suggest you make the calculations yourself with the WolframAlfa program. 

Chapter 11.8 Centers of gravity scn of the even square wave trajectory for n=0,1,2,… 8 and ω0=1/sec
Chapter 11.8.1 Introduction
We will repeat chaper 8, but this time we will calculate the centers of gravity of the trajectory scn using the WolframAlfa program. Previously, we took them on faith.

Fig. 11-23
Even square wave f(t) A=1, ω=1/sec, ϕ=0 and ω=50%.
We will calculate the successive harmonics with the WolframAlfa program using the formulas Fig. 11-1. But first we will check how WolframAlfa deals with square waves.

Chapter 11.8.2 Square wave f t) and WolframAlfa
We know what an instruction of a function looks like, e.g. sin(t) in WolframAlfa.
The function sin(t) is just sin(t) and the natural logarym ln(t) is ln(t). A square wave, on the other hand, is squarewave[t].
Let’s plot this function with the plot instruction.
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Type or paste into the box
plot squarewave[t] from t=-2 to 2

Fig.11-24
Squarewave[t] from t = -2 to 2
It is an odd, periodic function T = 1sec, A = 2 without a constant component. How to modify it to get the even as in Fig.11-23?
One should
1. Multiply by 0.5 to reduce the amplitude from A=2 to A=1
2. “Stretch” from T=1sec to T=2πsec
3. Move to the left π/2sec to change the odd function to even.
4. Move up 0.5
Let’s test it
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Type or paste into the box
plot 0.5*squarewave[(t+0.5pi)/(2pi)]+0.5 from t=-2π to 2π

Fig.11-25
The square wave from Fig. 11-23 generated by the WolframAlfa program.
The period T=2π is shown, more precisely from -π to + π.
In the following chapters we will designate its trajectories, centers of gravity scn and harmonics hn(t).

Chapter 11.8.3 Center of gravity sc0, that is for ω = 0, that is the constant component a0.
We will use the formula Fig. 11-1c
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Type or paste into the box
(1/(2pi))*integrate [[0.5+ 0.5 * SquareWave [(t + pi/2)/(2pi)]] from t=-pi to t=+pi

Fig.11-26
Calculation the constant component sc0=c0=a0=+0.5
As expected, a0=0.5 as the mean value of the square wave over the period -π to + π.
It was not even necessary to integrate, because a0 as the mean can be seen in Fig. 11-25.

Chapter 11.8.4 Center of gravity sc1, that is for ω=-1/sec.
So sc1 of the trajectory F (-1j1t)=f (t)exp(-1j1t) where f(t) is a square wave from Fig. 11-25.
We will repeat the animation from Fig. 8-3 chap. 8.

Fig. 11-27
Trajectory F (1j1t) of an even square wave
a. Vector 1exp(1j1t) as radius R=1 rotating at speed ω=-1/sec
b. Vector F(1j1t) as a radius R=1 modulated by a square wave f (t) from Fig. 11-25.
c. Trajectory F(1j1t) of an even square wave drawn by the vector b.
The vector of the center of gravity of the trajectory sc1=(+1 /π, 0) more or less agrees with the intuition, But as the mean of the rotating vector b must lie somewhere between (0,0) and (1,0).
Now let’s count it exactly with the WolframAlfa program. We will use the formula Fig. 11-1c.
Click https://www.wolframalpha.com. Type or paste into the box
[(1/(2pi)]*integrate[[0.5+0.5*SquareWave[(t+pi/2)/(2pi)]]*exp(-1i1t) dt from t=0 to t=+2pi

Fig. 1-28
Calculation of sc1=(1/ π, 0)
The program calculated the average of the vectors over time T=2πsec and it came out 0.31831.
First, it calculated a more precise value but rounded to 5 decimal places.
Second, it is exactly the number 1/π
Third, it is a vector in the form of a complex number sc1=(1/π, 0)

Chapter 11.8.5 Center of gravity sc2=0, that is for ω=-2/sec.
So sc2 of the trajectory F(-2j1t)=f (t)exp(-2j1t) where f (t) is a square wave from Fig. 11-23. We will repeat the animation from Fig. 8-5 chapter. 8.

Fig. 11-29
Trajectory F(2j1t) of an even square wave
a. Vector 1exp (j2j1t) as radius R=1 rotating at speed ω=-2/sec
b. Vector F(2j1t) as radius R=1 modulated by square wave f(t) from Fig. 11-25.
c. Trajectory F(2j1t) of an even square wave drawn by the vector b. The trajectory center of gravity vector sc2=(0,0) agrees as the mean.
Now let’s count it exactly with the WolframAlfa program.
Click https://www.wolframalpha.com. Type or paste into the box
[(1/(2pi)]*integrate [[0.5+ 0.5*SquareWave[(t+pi/2)/(2pi)]]*exp(-2i1t)dt from t=0 to t=+2pi

Fig. 11-30
Calculation sc2=0=(0,0)
That is, the second harmonic, more precisely for ω=2/sec, doesn’t exist

Chapter 11.8.6 Center of gravity sc3, that is for ω=-3/sec.
So sc3 trajectory F(-3j1t)=f (t)exp(-3j1t) where f (t) is a square wave from Fig. 11-25. We will repeat the animation from Fig. 8-6 chapter. 8.

Fig. 11-31
Trajectory F(3j1t) of an even square wave
a. Vector 1exp (j3j1t) as radius R=1 rotating at speed ω=-3/sec
b. Vector F(3j1t) as radius R=1 modulated by square wave f(t) from Fig. 11-25.
c. Trajectory F(3j1t) of an even square wave drawn by the vector b
Vector of the center of gravity of the trajectory sc3=(-1/3π, 0). The animation b is best suited for interpretation. Notice that the vector makes 2×3/4 turns. The left direction of the vector on the Re z axis is obvious when we consider 2 empty quarter turns. If they were not there (2×1 turn), then sc3=(0,0). And why is the length of the vector sc3 smaller than the length of sc1 in Figure 11-27c? Because there are more gaps in the period T=2πsec, which reduces the average. Ultimately, WolframAlfa will convince you.
We will use the formula Fig. 11-1c
Click https://www.wolframalpha.com.
Type or paste into the box
[(1/(2pi)]*integrate [[0.5+ 0.5*SquareWave[(t+pi/2)/(2pi)]]*exp(-3i1t)dt from t=0 to t=+2pi

Fig.1-32
Calculation sc3=(-1/3π,0)

Chapter 11.8.5 Center of gravity sc4=0, that is for ω=-4/sec.
So sc4 of the trajectory F(-4j1t)=f (t)exp (-4j1t) where f (t) is a square wave from Fig. 11-25. We will repeat the animation from Fig. 8-5 chapter. 8.

Fig. 11-33
Trajectory F(4j1t) of an even square wave
a. Vector 1exp(4j1t) as radius R=1 rotating at speed ω=-4/sec
b. Vector F(4j1t) as radius R=1 modulated by square wave f (t) from Fig. 11-25.
c. Trajectory F(4j1t) drawn by the vector from b.
The trajectory center of gravity vector sc4=(0,0) agrees as the mean.
Now let’s count it exactly with the WolframAlfa program. We will use the formula Fig. 11-1c
We will use the formula Fig. 11-1c
Click https://www.wolframalpha.com.
Type or paste into the box
[(1/(2pi)]*integrate [[0.5+ 0.5*SquareWave[(t+pi/2)/(2pi)]]*exp(-4i1t)dt from t=0 to t=+2pi

Fig. 11-34
Calculation sc4=0=(0,0)
That is, the fourth harmonic, more precisely for ω=4/sec, doesn’t exist

Chapter 11.8.8 Centers of gravity sc5, sc6, sc7 and sc8, that is for ω = -5 / sec, -6 / sec, -7 / sec and- 8 / sec.
As homework. I will just suggest that it is enough to paste it into the WolframAlfa window [(1/(2pi)]*integrate [[0.5+ 0.5*SquareWave[(t+pi/2)/(2pi)]]*exp(-ni1t)dt from t=0 to t=+2pi
with an appropriately changed parameter n.