Math Is Fun Forum

gnitsuk ,you are a better teacher than me, i should say.

moreover, i want to say a few words to RickyGeorge

You don't need to be confused by the arguement by the two parts of your id

Actually, mathematicians have carefully chosen a defination to avoid the "approaching vs being" controversy.

They define an indefinite integral as the limit that a definate integral approaches. so the same symbol may mean different things with regard to whether an infinity or an undefined point is involved.You can consult to your caculus book and check out if these two types of integrals are  in an advanced chapter.

whether being or approaching is not that important if you accept a consept that math itself is a theory, which is an approximition. As a philosopher once put it "You cannot find two idendical leaves" --1+1=2 is nonetheless, an approximation, too

∫ e^(tx) / β  e^(-x/β)  dx
do you mean it by ∫e[sup]tx/β[/sup]e[sup]-x/β[/sup] dx ?
or ∫e[sup]tx[/sup]/(β e[sup]-x/β[/sup]) dx?

well, i may 1+1 in some way.

but 0.111... can be explained as infinite numbers adding together (different from varying variable), if you don't insist on interpreting it as drawings of 1s

i'm not good at integration and i'm not sure whether i did it correct or not

1st tech

since you wanna find its integral, or antiderivative, F(x) is one of the answers if its derivative is truely the one whose derivative you wanna find

2nd tech

F(x)+Cs  share the same derivative with F(x), F(x)+C is also the antiderivative of F'(x)

3rd tech

given another one function satisfying G'(x)=F'(x) G(x) must be of the form F(x)+C

for (G(x)-F(x))'=G'(x)-F'(x)=0 and only C'=0

Conclusion: once you've found a function F(x) whose derivative is the one you wanna integrate, F(x)+c represents any its antiderivative

4th tech

(Cf(x))'=C(f(x))' or=Cf'(x) where C is a constant

5th tech

f'(x)=dy/dx x is any variable defined in domain

hence x itself could be a function too, as long as its value vary in the domain

x=g(z) g'(z)=dx/dz

it makes sense that

dy/dx*dx/dz=dy/dz

hence dy/dz= (f(g(z)))' of z=dy/dx dx/dz= f'(x) g'(z)=f'(g(z)) g'(z)

substitude x for z in the formula above

using tech 3 tech4 and tech5, do you understand how to solve problem 1 now?

because in order to prove Euler's formula, we commonly use Taylor series.

And both Sine and Cosine Taylor series are derived via making derivatives (i don't know how to use a single verb to say it) of themselves.

Sine's derivative's proof has an essential part:

Sin(x+Δx) - Sin(x) = 2Sin((x+Δx-x)/2)Cos((x+Δx+x)/2)=2Sin(Δx/2)Cos(x+Δx/2)

More or less does cosine's

0.111...123 is a number too, uh??

Circular proof, i am sorry to say

Why don't you explore other applications, to which cannot be solved simply within trigonometric formulas?

eg.

Sin[A] Sin[b] Sin[C]

(Ns - N) / (Ns)=Ns/Ns-N/(Ns)=1-1/s

As long as S=1-1/s, no matter what N is ,the equation S = (Ns - N) / (Ns)
is valid. That means N is Undetermined by this equation! (N can be any number)

another example a+2N=a+N+N, can you solve N out?

Yeah, as Mr Hollis put in his Caculus book, we are using sum of infinite series IMPLICITLY.

Numbers are a theory, "theory" means approximation

0.99... is accurate to leave perhaps only one particle uncollected from recursively cutting a cake into equally 10 pieces, then collecting 9 and leaving the other one to be next round cut-ee.

Suggest you to get a Derive software

i am bored with integration solving, haha

1) [e^(-x/β)]'=e^(-x/β) (-x/β)'= - 1/β e^(-x/β)

∫ 1/β e^(-x/β)  dx = -e^(-x/β)+C

2) ∫ x/β e^(-x/β)  dx =  ∫xd(-e^(-x/β)) = -∫xd(e^(-x/β))= -xe^(-x/β)+∫e^(-x/β)dx =  -xe^(-x/β) - βe^(-x/β)+ C

i am tired...

No I'm not.  It's goes on infinitely long, and thus is never ending.  You can never reach then end. ----------

Thus i am not sure if the result 0.1 0.11 0.111 0.11...1 (k 1's), =>0.11...11 (k+1 1's) gotten from mathematical induction could catch up your number, for digits of your number never ends, my never ends, either.

Similar to your argument, infinity-infinity is an indeterminate form i don't know if  my infinite 1's could equal yours.

i don't know how to compare with my  step by step growing number with your existing never ending "number", mathematical induction doesn't garantee this, what it does say is  divided to a finite digit, it can be divided to 1 more digit.

if your question is (3/5+i4/5)[sup]100[/sup]=?the solution is words below

Key theories:

(1) (cosa+i sina)(cosb+i sinb)
= cosa cosb - sina sinb + (sina cosb + cosa sinb)i

= cos(a+b)+ sin(a+b)i

(cosa+i sina)(cosa+i sina)(cosa+i sina) = (cos2a+isin2a)(cosa+i sina) =cos3a+isin3a

Similarly, (cosx+i sinx)[sup]k[/sup]=cos(kx)+i sin(kx)......(1b)

cos(x)²+sin(x)²=1    (3)

any a+bi can be expressed as
a+bi = (a²+b²)(a+bi)/(a²+b²) = (a²+b²) (a/(a²+b²)+b/(a²+b²) i)
a/(a²+b²) and b/(a²+b²) satisfy cos(x) and sin(x) in equation(3) . Actually we need  not point out angle x explicitly sometimes, besides, i forgot the formula!

(a+bi)[sup]k[/sup]= (a²+b²)[sup]k[/sup](...)  (4)

As for this case,

(3/5)²+(4/5)²=1 angle x could be arcsin(3/5)

the answer is cos[100arcsin(3/5)]+i sin[100arcsin(3/5)]

As far as i can reach