H E A Lt H Y P R E G N A N C Y: Fill & Download for Free

Check The International Obfuscated C Code Contest (IOCCC) WinnersMy favs:1:main(_){_^448&&main(-~_);putchar(--_%64?32|-~7[__TIME__-_/8%8][">'txiZ^(~z?"-48]>>";;;====~$::199"[_*2&8|_/64]/(_&2?1:8)%8&1:10);} 2:#include <math.h> #include <sys/time.h> #include <X11/Xlib.h> #include <X11/keysym.h>  double L ,o ,P  ,_=dt,T,Z,D=1,d,  s[999],E,h= 8,I,  J,K,w[999],M,m,O  ,n[999],j=33e-3,i=  1E3,r,t, u,v ,W,S=  74.5,l=221,X=7.26,  a,B,A=32.2,c, F,H;  int N,q, C, y,p,U;  Window z; char f[52]  ; GC k; main(){ Display*e=  XOpenDisplay( 0); z=RootWindow(e,0); for (XSetForeground(e,k=XCreateGC (e,z,0,0),BlackPixel(e,0)) ; scanf("%lf%lf%lf",y +n,w+y, y+s)+1; y ++); XSelectInput(e,z= XCreateSimpleWindow(e,z,0,0,400,400, 0,0,WhitePixel(e,0) ),KeyPressMask); for(XMapWindow(e,z); ; T=sin(O)){ struct timeval G={ 0,dt*1e6} ; K= cos(j); N=1e4; M+= H*_; Z=D*K; F+=_*P; r=E*K; W=cos( O); m=K*W; H=K*T; O+=D*_*F/ K+d/K*E*_; B= sin(j); a=B*T*D-E*W; XClearWindow(e,z); t=T*E+ D*B*W; j+=d*_*D-_*F*E; P=W*E*B-T*D; for (o+=(I=D*W+E *T*B,E*d/K *B+v+B/K*F*D)*_; p<y; ){ T=p[s]+i; E=c-p[w]; D=n[p]-L; K=D*m-B*T-H*E; if(p [n]+w[ p]+p[s ]== 0|K <fabs(W=T*r-I*E +D*P) |fabs(D=t *D+Z *T-a *E)> K)N=1e4; else{ q=W/K *4E2+2e2; C= 2E2+4e2/ K  *D; N-1E4&& XDrawLine(e ,z,k,N ,U,q,C); N=q; U=C; } ++p; } L+=_* (X*t +P*M+m*l); T=X*X+ l*l+M *M;  XDrawString(e,z,k ,20,380,f,17); D=v/l*15; i+=(B *l-M*r -X*Z)*_; for(; XPending(e); u *=CS!=N){  XEvent z; XNextEvent(e ,&z);  ++*((N=XLookupKeysym  (&z.xkey,0))-IT?  N-LT? UP-N?& E:&  J:& u: &h); --*(  DN -N? N-DT ?N==  RT?&u: & W:&h:&J  ); } m=15*F/l;  c+=(I=M/ l,l*H  +I*M+a*X)*_; H  =A*r+v*X-F*l+(  E=.1+X*4.9/l,t  =T*m/32-I*T/24  )/S; K=F*M+(  h* 1e4/l-(T+  E*5*T*E)/3e2  )/S-X*d-B*A;  a=2.63 /l*d;  X+=( d*l-T/S  *(.19*E +a  *.64+J/1e3  )-M* v +A*  Z)*_; l +=  K *_; W=d;  sprintf(f,  "%5d %3d"  "%7d",p =l  /1.7,(C=9E3+  O*57.3)%0550,(int)i); d+=T*(.45-14/l*  X-a*130-J* .14)*_/125e2+F*_*v; P=(T*(47  *I-m* 52+E*94 *D-t*.38+u*.21*E) /1e2+W*  179*v)/2312; select(p=0,0,0,0,&G); v-=(  W*F-T*(.63*m-I*.086+m*E*19-D*25-.11*u  )/107e2)*_; D=cos(o); E=sin(o); } } There is also a tool, Mangle-It C++ Obfuscator, which makes C/C++ code unreadable in order to guard your intellectual property and defeat hackers.

Absolutely necessary! IOCCC Flight Simulator includes build instructions.#include <math.h> #include <sys/time.h> #include <X11/Xlib.h> #include <X11/keysym.h>  double L ,o ,P  ,_=dt,T,Z,D=1,d,  s[999],E,h= 8,I,  J,K,w[999],M,m,O  ,n[999],j=33e-3,i=  1E3,r,t, u,v ,W,S=  74.5,l=221,X=7.26,  a,B,A=32.2,c, F,H;  int N,q, C, y,p,U;  Window z; char f[52]  ; GC k; main(){ Display*e=  XOpenDisplay( 0); z=RootWindow(e,0); for (XSetForeground(e,k=XCreateGC (e,z,0,0),BlackPixel(e,0)) ; scanf("%lf%lf%lf",y +n,w+y, y+s)+1; y ++); XSelectInput(e,z= XCreateSimpleWindow(e,z,0,0,400,400, 0,0,WhitePixel(e,0) ),KeyPressMask); for(XMapWindow(e,z); ; T=sin(O)){ struct timeval G={ 0,dt*1e6} ; K= cos(j); N=1e4; M+= H*_; Z=D*K; F+=_*P; r=E*K; W=cos( O); m=K*W; H=K*T; O+=D*_*F/ K+d/K*E*_; B= sin(j); a=B*T*D-E*W; XClearWindow(e,z); t=T*E+ D*B*W; j+=d*_*D-_*F*E; P=W*E*B-T*D; for (o+=(I=D*W+E *T*B,E*d/K *B+v+B/K*F*D)*_; p<y; ){ T=p[s]+i; E=c-p[w]; D=n[p]-L; K=D*m-B*T-H*E; if(p [n]+w[ p]+p[s ]== 0|K <fabs(W=T*r-I*E +D*P) |fabs(D=t *D+Z *T-a *E)> K)N=1e4; else{ q=W/K *4E2+2e2; C= 2E2+4e2/ K  *D; N-1E4&& XDrawLine(e ,z,k,N ,U,q,C); N=q; U=C; } ++p; } L+=_* (X*t +P*M+m*l); T=X*X+ l*l+M *M;  XDrawString(e,z,k ,20,380,f,17); D=v/l*15; i+=(B *l-M*r -X*Z)*_; for(; XPending(e); u *=CS!=N){  XEvent z; XNextEvent(e ,&z);  ++*((N=XLookupKeysym  (&z.xkey,0))-IT?  N-LT? UP-N?& E:&  J:& u: &h); --*(  DN -N? N-DT ?N==  RT?&u: & W:&h:&J  ); } m=15*F/l;  c+=(I=M/ l,l*H  +I*M+a*X)*_; H  =A*r+v*X-F*l+(  E=.1+X*4.9/l,t  =T*m/32-I*T/24  )/S; K=F*M+(  h* 1e4/l-(T+  E*5*T*E)/3e2  )/S-X*d-B*A;  a=2.63 /l*d;  X+=( d*l-T/S  *(.19*E +a  *.64+J/1e3  )-M* v +A*  Z)*_; l +=  K *_; W=d;  sprintf(f,  "%5d %3d"  "%7d",p =l  /1.7,(C=9E3+  O*57.3)%0550,(int)i); d+=T*(.45-14/l*  X-a*130-J* .14)*_/125e2+F*_*v; P=(T*(47  *I-m* 52+E*94 *D-t*.38+u*.21*E) /1e2+W*  179*v)/2312; select(p=0,0,0,0,&G); v-=(  W*F-T*(.63*m-I*.086+m*E*19-D*25-.11*u  )/107e2)*_; D=cos(o); E=sin(o); } } 

Okay, let’s prove this thing from start to finish. As long as you understand what a function is and a little bit of algebra, you should be able to understand this whole thing. I’ve divided it into 13 subsections so you can skip what you know. I also reference multiple figures; those are graphics I found on the internet and put at the bottom because graphics and lists don’t play nice in Quora answers.Logarithms and properties of logarithms.First, a logarithm answers the question “What power to I have to raise a number to in order to get another number.”[math]log_{a}(b)=c[/math] is pronounced “log base a of b equals c”, and means [math]a^c=b.[/math] An example of a true logarithm is [math]log_{2}(8)=3,[/math] since [math]2^3=8.[/math]The notation [math]log(x)[/math] is usually equivalent to [math]log_{10}(x).[/math] The notation [math]ln(x)[/math] is pronounced “natural log of x”, and is equivalent to [math]log_{e}(x)[/math], where [math]e[/math] is an irrational constant approximately equal to [math]2.72.[/math] I further explain the number in section 3.Knowing that [math]a^{q+r}=a^q \times a^r[/math], substitute [math]log_{a}(b)[/math] in for [math]q[/math] and [math]log_{a}(c)[/math] in for [math]r[/math], you get [math]a^{log_{a}(b)+log_{a}(c)}=a^{log_{a}(b)} \times a^{log_{a}(c)}.[/math] The [math]a[/math] in base on the RHS cancels with the [math]log_{a}[/math], yielding [math]a^{log_{a}(b)+log_{a}(c)}=b \times c.[/math]Take [math]log_{a}[/math] of both sides. That cancels out the [math]a[/math] base on the LHS, yielding [math]log_{a}(b)+log_{a}(c)=log_{a}(b \times c).[/math] Thus proves that the sum of logs is equal to the log of a product.Using similar techniques with [math](a^q)^r=a^{q \times r}[/math] and [math]a^{q-r}=\frac{a^q}{a^r}[/math], one can show that [math]c \times log_{a}(b)=log_{a}(b^c) [/math] and [math]log_{a}(b)-log_{a}(c)=log_{a}(\frac{b}{c}).[/math] These formulae will be important in step 5.Right triangles, the Pythagorean Theorem, and its proof.First, a right triangle is a triangle with one right angle (corner). A right angle is the angle in a square.A right triangle has two legs- sides adjacent to the right angle- and a hypotenuse- the side that’s not a leg.The square of the lengths of the legs is always equal to the square of the length of the hypotenuse, normally written as [math]a^2+b^2=c^2.[/math] This is a statement called the Pythagorean Theorem.To prove the Pythagorean Theorem true, I will use Figure A at the bottom of this answer.Notice that there are 4 right triangles, with base and height [math]a[/math] and [math]b[/math] and an inside square with sides [math]c[/math].The area of a triangle is [math]\frac{1}{2} \times base \times height[/math] and the area of a square is [math]side^2.[/math] Thus, the area of the triangles plus the inside square is [math]4 \times \frac{1}{2} a b + c^2.[/math]The triangles and inside square also comprise the outside square, which has the area [math](a+b)^2.[/math] Since those areas are equal, [math](a+b)^2=4 \times \frac{1}{2} a b + c^2.[/math]Distribute the left and simplify the right: [math]a^2+2ab+b^2=2ab + c^2.[/math]Subtract [math]2ab[/math] from both sides: [math]a^2+b^2=c^2.[/math]Thus proves the Pythagorean TheoremSine and cosine, their additive formulas, and Pythagorean identities.The graphic in Figure B explains sine and cosine better than me.Importantly, the values of sine, cosine, and tangent are invariant to the size of the triangle. Knowing only the size of the angle, you can immediately know the sine, cosine, and tangent.The definitions of sine, cosine, and tangent have been extended to all real numbers, including negative angles and angles bigger than a right angle, by imagining negative length sides. Using the measure of degrees, such that a right angle is [math]90^o[/math], the following are the rules for extending beyond the range of a right triangle:[math]x=0: sin(x)=0, cos(x)=1[/math][math]x=90: sin(x)=1, cos(x)=0[/math][math]90 \lt x \lt 180: sin(x)=sin(180-x), cos(x)=-cos(180-x)[/math][math]180 \le x \le 270: sin(x)=-sin(x-180), cos(x)=-cos(x-180)[/math][math]270 \lt x \lt 360: sin(x)=-sin(360-x), cos(x)=cos(360-x)[/math][math]x \ge 360:[/math] Subtract [math]360[/math] from [math]x[/math] repeatedly until [math]0 \le x \lt 360[/math] and treat it as the result of that. For example, [math]sin(412)=sin(52).[/math][math]x \lt 0:[/math] Add [math]360[/math] to [math]x[/math] repeatedly until it [math]0 \le x \lt 360[/math] and treat it as the result of that. For example, [math]sin(-308)=sin(52).[/math]See 3f for the value of tangent at an arbitrary real number.The graphic in Figure C shows the proof for some of the formulas you’ll need. The ones you’ll need later are [math]sin(x+y)=sin(x)cos(y)+cos(x)sin(y)[/math] and [math]cos(x+y)=cos(x)cos(y)-sin(x)sin(y):[/math] look for each of those in the graphic to understand. They’re referred to as the additive properties of sine and cosine.The formula from section 2 states that [math]a^2+b^2=c^2.[/math] Select one angle, and the formula becomes [math]opp^2+adj^2=hyp^2[/math] Divide both sides by [math]hyp^2 [/math]yields [math]opp^2/hyp^2+adj^2/hyp^2=1.[/math] Substituting sine and cosine at their respective places yields [math]sin^2(x)+cos^2(x)=1.[/math][math]tan(x)[/math] is always equal [math]\frac{sin(x)}{cos(x)}[/math]. This is because the fraction comes out to [math]\frac{\frac{opp}{hyp}}{\frac{adj}{hyp}}.[/math] The hypotenuses cancel out, yielding [math]\frac{opp}{adj},[/math] which is the definition of [math]tan(x).[/math]Limits and the number [math]e[/math] explained.Limits are written as [math]\lim_{n \to c} f(n).[/math] The value of the expression answers “What does the value of [math]f(n)[/math] approach as [math]n[/math] gets arbitrarily close to [math]c[/math]?”They have a rigorous definition, but I’m not going to get into it.Most of the time, the expression is simply equal to [math]f(c)[/math].On occasion, however, they are useful outside of that. An example of a function where they might be useful is [math]\lim_{x \to 0} \frac{x^2}{x}.[/math] The function is not defined at 0, since that would see a 0 in the denominator. However, the limit lets you treat [math]x[/math] as “a value arbitrarily close to but not equal to 0”, which lets you cancel [math]x[/math] in the numerator and denominator, yielding [math]\lim_{x \to 0} x,[/math] which approaches 0. While this example may seem contrived, there are times where it makes sense, which I will explore later.Finally, you will sometimes see [math]\lim_{x \to \infty}[/math], which means “as x gets arbitrarily large.”For explaining the number [math]e[/math], it is defined as [math]lim_{n \to \infty}(1 + \frac{1}{n})^n.[/math]Derivatives explained.The derivative of a function [math]f(x)[/math] can be notated as [math]f'(x)[/math] or [math]\frac{df(x)}{dx}.[/math][math]f'(x)[/math] tells the value of the slope at any point on [math]f(x).[/math]The slope between two points is rise over run. For example, the slope between [math](2,4)[/math] and [math](1,1)[/math] is [math](4-1)/(2-1)=3/1=3.[/math]So to calculate the slope of a function at one point, we need that point, and another to bring arbitrarily close to the first point. From there we’ll calculate rise over run. As such, the formula is: [math]lim_{h \to 0} \frac{f(x+h)-f(x)}{(x+h)-x}[/math] or [math]lim_{h \to 0} \frac{f(x+h)-f(x)}{h}.[/math] This formula is called the “definition of derivative.”The nth derivative, notated as [math]f^{(n)}(x)[/math] (the parentheses on the [math]n[/math] are necessary) or [math]\frac{d^nf(x)}{dx^n}[/math] means the derivative of the (n-1)th derivative, where the 0th derivative is the original function. So the 1st derivative, [math]f^{(1)}(x),[/math] is the derivative [math]f'(x)[/math]. The 2nd derivative, [math]f^{(2)}(x),[/math] is the derivative of the 1st derivative, [math]f^{(1)}{'}(x)[/math] or [math]f''(x)[/math]. The 3rd is the derivative of the 2nd, and so on.Proof that [math]\frac{d\ln(x)}{dx}=\frac{1}{x}.[/math]Using the definition of derivative, [math]\frac{d\ln(x)}{dx}=lim_{h \to 0} \frac{ln(x+h)-ln(x)}{h}.[/math]Put the [math]\frac{1}{h}[/math] in front: [math]lim_{h \to 0} \frac{1}{h} \times (ln(x+h)-ln(x)).[/math]Use log properties from section 1: [math]lim_{h \to 0} ln(\frac{x+h}{x}^{\frac{1}{h}}).[/math]Separate the fraction in the logarithm: [math]lim_{h \to 0} ln((1+\frac{h}{x})^{\frac{1}{h}}).[/math]Set [math]q[/math] to [math]1/h[/math]. As [math]h[/math] approaches 0, [math]q[/math] approaches [math]\infty[/math]. [math]lim_{q \to \infty} ln((1+\frac{1}{qx})^q).[/math]Add a [math]\frac{x}{x}[/math] to the beginning. Use a log property from section 1 to move the [math]x[/math] in the numerator to the exponent in the log. [math]lim_{q \to \infty} \frac{1}{x} \times ln((1+\frac{1}{qx})^{qx}).[/math]As [math]q[/math] approaches [math]\infty[/math], so does [math]qx.[/math] So the part inside the logarithm fits the definition of [math]e[/math] from section 5. That gives us [math]\frac{1}{x} \times ln(e).[/math]The expression [math]ln(e)[/math] means “e to what power is e?” Since they’re the same number, the answer to that is 1. This leaves [math]\frac{1}{x} \times 1,[/math] or [math]\frac{1}{x}[/math].Thus, [math]\frac{d\ln(x)}{dx}=\frac{1}{x}.[/math]Proof that [math]\frac{df(g(x))}{dx}=f'(g(x)) \times g'(x)[/math] (chain rule).Using the definition of derivative, [math]\frac{df(g((x))}{dx}=lim_{h \to 0} \frac{f(g(x+h))-f(g(x))}{h}.[/math]Multiply by [math]\frac{g(x+h)-g(x)}{g(x+h)-g(x)}[/math] and switch the denominators: [math]lim_{h \to 0} \frac{f(g(x+h))-f(g(x))}{g(x+h)-g(x)} \times \frac{g(x+h)-g(x)}{h}.[/math]The first expression is the same rise over run showed in section 5, only with [math]g(x)[/math] as the run. The second expression shows the definition of derivative for [math]g(x).[/math]Thus, [math]\frac{df(g(x))}{dx}=f'(g(x)) \times g'(x)[/math]Proof that [math]\frac{d e^x}{dx}=e^x.[/math][math]f(x)=e^x[/math]Take the natural log of both sides: [math]ln(f(x))=x[/math]Take the derivative of both sides with respect to x, using the derivative in section 6 and the chain rule from section 7: [math]\frac{f'(x)}{f(x)}=1.[/math]Multiply both sides by [math]f(x): f'(x)=f(x).[/math]Thus, the derivative of [math]e^x[/math] is itself. In other words, [math]\frac{d e^x}{dx}=e^x.[/math]Proof that [math]\frac{d sin(x)}{dx}=q \times cos(x),[/math] [math]\frac{d cos(x)}{dx}=-q \times sin(x),[/math] and [math]\frac{d (c \times f(x))}{dx}=c \times \frac{d f(x)}{dx},[/math] establishing a cycle.For sine, by the definition of derivative, [math]lim_{h \to 0} \frac{sin(x+h)-sin(x)}{h}.[/math]Using the formulas in section 3, [math]lim_{h \to 0} \frac{sin(x)cos(h)+cos(x)sin(h)-sin(x)}{h}.[/math]Separate different parts of the fraction and factor out a [math]sin(x)[/math]: [math]lim_{h \to 0} \frac{sin(x)(cos(h)-1)}{h} + \frac{cos(x)sin(h)}{h}.[/math]Multiply the left side by [math]\frac{1+cos(h)}{1+cos(h)},[/math] and invert [math]cos(h)-1[/math] by bringing out a negative sign: [math]lim_{h \to 0} -\frac{sin(x)(1+cos(h))(1-cos(h))}{h(1+cos(h))} + \frac{cos(x)sin(h)}{h}.[/math]Multiply the parenthetical expressions on the left side: [math]lim_{h \to 0} -\frac{sin(x)(1-cos^2(h))}{h(1+cos(h))} + \frac{cos(x)sin(h)}{h}.[/math]Use the identity in section 3 to simplify the left size: [math]lim_{h \to 0} -\frac{sin(x)(sin^2(h))}{h(1+cos(h))} + \frac{cos(x)sin(h)}{h}.[/math]Separate the fractions into solveable chunks: [math]lim_{h \to 0} -sin(x) \times \frac{sin(h)}{h} \times \frac{sin(h)}{(1+cos(h))} + cos(x) \times \frac{sin(h)}{h}.[/math]We cannot solve the value of [math]lim_{h \to 0} \frac{sin(h)}{h}[/math] yet. It depends on the units used for sine and cosine. For now, we’ll use the value [math]q: -sin(x) \times q \times \frac{sin(0)}{(1+cos(0))} + cos(x) \times q.[/math]Simplify further: [math]-sin(x) \times q \times \frac{0}{2} + cos(x) \times q.[/math]Simplify again: [math]q\cos(x).[/math]Next, with cosine. By the definition of derivative, [math]lim_{h \to 0} \frac{cos(x+h)-cos(x)}{h}.[/math]Using the identity in section 3: [math]lim_{h \to 0} \frac{cos(x)cos(h)-sin(x)sin(h)-cos(x)}{h}.[/math]Separate the fraction and factor out a [math]cos(x)[/math]: [math]lim_{h \to 0} \frac{cos(x)(cos(h)-1)}{h}-\frac{sin(x)sin(h)}{h}.[/math]Using the same steps as 9d-9f, simplify the left side: [math]lim_{h \to 0} -\frac{cos(x)(sin^2(h))}{h(cos(h)+1)}-\frac{sin(x)sin(h)}{h}.[/math]Separate the fractions similar to in 9g: [math]lim_{h \to 0} -cos(x) \times \frac{sin(h)}{h} \times \frac{sin(h)}{(1+cos(h))} - sin(x) \times \frac{sin(h)}{h}.[/math]Simplify: [math]-cos(x) \times q \times \frac{sin(0)}{(1+cos(0))} - sin(x) \times q.[/math]Simplify further: [math]-q\sin(x)[/math]Finally, for [math]c f(x).[/math] By the definition of derivative, [math]lim_{h \to 0} \frac{c f(x-h) - c f(x)}{h}.[/math]Factor out c: [math]c \times lim_{h \to 0} \frac{f(x-h) - f(x)}{h}.[/math]Now that we’ve established that constants on a function can be applied to the derivative, it follows that [math]\frac{d(-sin(x))}{dx}[/math] is [math]-q\cos(x)[/math] and [math]\frac{d(-cos(x))}{dx}[/math] is [math]q\sin(x).[/math] Thus, repeatedly taking higher order derivatives of sine and cosine yields the cycle [math]sin, cos, -sin, -cos[/math] with increasing powers of [math]q.[/math]Proof that the value of [math]q[/math] from section 9 is equal to [math]1[/math] if and only if sine and cosine are measured in radians.First, draw a unit circle and lines with points labelled according to Figure D.Using the area of a triangle formula from section 2, the area of the triangle that’s brown is [math]\frac{1}{2} \times 1 \times sin(\theta).[/math]The area of the triangle that’s brown, green, and blue is [math]\frac{1}{2} \times 1 \times tan(\theta).[/math]The sector that’s brown and green is a part of a circle. This is the part where we determine the units for sine and cosine. I’ll use the units of radians, which means [math]2\pi[/math] is a full circle.The area of a circle is [math]\pi r^2.[/math] So the area of a sector spanning one radian is [math]\frac{\pi r^2}{2\pi}=\frac{r^2}{2}.[/math] In this case, the circle is of radius [math]1[/math], so the angle is [math]\frac{1}{2} \times \theta[/math], where [math]\theta[/math] is the amount of radians the angle spans.We can see that the triangle from 10c contains the sector from 10e, which contains the triangle from 10b. Thus, the containing shape’s area must be at least as large as that of the contained shape: [math]\frac{tan(\theta)}{2} \ge \frac{\theta}{2} \ge \frac{sin(\theta)}{2}.[/math]Multiply all terms by [math]2[/math] and represent all trig terms as sine and cosine: [math]\frac{sin(\theta)}{cos(\theta)} \ge \theta \ge sin(\theta).[/math]Divide all terms by [math]sin(\theta): \frac{1}{cos(\theta)} \ge \frac{\theta}{sin(\theta)} \ge 1.[/math]Now, apply [math]lim_{\theta \to 0}.[/math] Since the middle was the reciprocal of [math]q[/math], it evaluates to [math]\frac{1}{q}[/math]:[math] \frac{1}{cos(0)} \ge \frac{1}{q} \ge 1.[/math]Simplify: [math]1 \ge \frac{1}{q} \ge 1.[/math]Since [math]\frac{1}{q}[/math] is both [math]\ge 1[/math] and [math]\le 1[/math], it must be equal to [math]1.[/math] Thus, [math]q=\frac{1}{1}=1.[/math]Since [math]q=1[/math] when dealing in radians, the increasing powers of [math]q[/math] mentioned in section 9 don’t matter. In radians, the derivatives [math]sin(x), cos(x), -sin(x), -cos(x)[/math] form a simple cycle.Taylor Series explained and the Taylor Series of [math]sin(x), cos(x),[/math] and [math]e^x.[/math]The Taylor Series of a function is [math]\sum_{n=0}^{\infty} \frac{f^{(n)}(a)(x-a)^n}{n!}[/math], where [math]n![/math] denotes the factorial function and any real number works for [math]a[/math]. Most times, people use [math]a=0[/math].[math]\sum_{n=p}^{q}[/math] means “evaluate the value of the following expression where [math]n=p[/math] and call that [math]r[/math]. Then, repeat for [math]n=p+1[/math] and add the result of that to [math]r[/math]. Repeat this process with every subsequent integer up to and including [math]n=q[/math], adding each subsequent result to the growing value of [math]r[/math].” Using the variable [math]r[/math] doesn’t matter, but it’s easier to explain using a variable there.The factorial function is as follows: [math]0!=1[/math], and for any integer higher than [math]0[/math], [math]n!=n\times(n-1)!.[/math] So [math]5!=5 \times 4 \times 3 \times 2 \times 1.[/math]The Taylor Series, when it converges, converges to the value of [math]f(x).[/math] Since the summation keeps going to [math]\infty[/math], the exact calculation cannot be finished. However, using more terms gets one closer to the result.The derivative of [math]e^x[/math] is itself. So all values [math]f^{(n)}(x)[/math] will be the same. At [math]x=0[/math], which is the value of [math]a[/math] I will be using, that comes out to [math]1.[/math]Thus, the Taylor Series from [math]e^x[/math] is [math]\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}...[/math] If you noticed the [math]x^0[/math] in the first term and are wondering about when [math]x=0[/math], when evaluating a Taylor Series, [math]x^0[/math] will evaluate to [math]1,[/math] even when [math]x=0.[/math]From the identities in section 3, the values of [math]sin(0), cos(0), -sin(0), and -cos(0)[/math] are respectively [math]0, 1, 0, -1[/math].Thus, the Taylor Series for [math]sin(x)[/math] is [math]\frac{x^1}{1!}-\frac{x^3}{3!}...[/math] and the Taylor Series for [math]cos(x)[/math] is [math]\frac{x^0}{0!}-\frac{x^2}{2!}...[/math] where x is measured in radians.The number [math]i[/math] explained and the result when [math]ix[/math] is put into the Taylor Series for [math]e^x.[/math]The value [math]-1[/math] does not have a real square root. This is because, in multiplication, a positive times a positive or a negative times a negative both yield positive numbers. Thus, no number when multiplied by itself is equal to [math]-1.[/math]The value [math]i[/math] was invented as an imaginary number to serve as [math]\sqrt{-1}.[/math] Since its inception, it has been used to turn the number line into a complex plane (google it for a visual), with all complex numbers represented as [math]a+bi.[/math] Examples include [math]3+4i, 2.7i, 1.2-6i,[/math] etc.[math]i[/math] does have a square root: [math]\frac{±(1+i)}{\sqrt{2}}.[/math] It even has a 4th root, an 8th root, and so on, but this answer will not get into that.The powers of [math]i[/math] form a cycle like the powers of [math]-1.[/math] [math]i^2=-1, i^3=-i, i^4=1, i^5=i,[/math] etc.When the value [math]ix[/math] is put into the Taylor Series for [math]e^x[/math] from section 11, the result is [math]\frac{(ix)^0}{0!}+\frac{(ix)^1}{1!}+\frac{(ix)^2}{2!}+\frac{(ix)^3}{3!}...[/math]Taking all [math]i[/math]'s out of their respective fractions and evaluating [math]i[/math] to each power: [math]\frac{(x)^0}{0!}+i\frac{(x)^1}{1!}-\frac{(x)^2}{2!}-i\frac{(x)^3}{3!}...[/math]Separate the parts with a remaining [math]i[/math] from those without: [math](\frac{(x)^0}{0!}-\frac{(x)^2}{2!}...)+(i\frac{(x)^1}{1!}-i\frac{(x)^3}{3!}...)[/math]Factor out [math]i[/math] from the right side: [math](\frac{(x)^0}{0!}-\frac{(x)^2}{2!}...)+i(\frac{(x)^1}{1!}-\frac{(x)^3}{3!}...)[/math]The part on the left is the Taylor Series for [math]cos(x)[/math] from section 11. The part on the right after factoring out [math]i[/math] is the Taylor Series for [math]sin(x)[/math] from section 11. Thus, [math]e^{ix}=\cos(x)+i\sin(x).[/math]The result of putting [math]\pi[/math] into the formula from section 12, and some algebra to finish the proof.[math]e^{i\pi}=\cos(\pi)+i\sin(\pi).[/math] Remember that unlike how I did section 3, this is in radians. In degrees, a straight angle is [math]180^o.[/math] In radians, that’s [math]\pi.[/math]Substituting the values for a straight angle: [math]e^{i\pi}=-1+0i=-1.[/math]Add [math]1[/math] to both sides: [math]e^{i\pi}+1=0.[/math]And that’s how, with a lot of math, and an anti-climactic end in algebra, you show that [math]e^{i\pi}+1=0.[/math]Figure A:Figure B:Figure C:Figure D: