Just find a nice demonstration for volumes and area of the n-spheres here.
First he demonstrates the formula for a general pyramid (by dividing a general cube in pyramids) to get the Volume from the Area. The case for the 3 dimensional sphere he settles by comparing its area with the cylinder, actually he just claims it is possible, pointing to a more famous argument that I already saw before here for example. Then he claims that this kind of argument always works, that I didn't knew!! And derives the general formula from it. It is actually a recurrence formula but we can certainly find the general formula from that.
A bizarre but nice fact of polynomials
The tangent to the midpoint between two roots in a third degree polynomial(with three roots) is always the third root. Just saw that at Polymath.
Deriving f(x)=K*(x-a)*(x-b)*(x-c) using the product rule we get
f'(x)= K*((x-a)*(x-b)+(x-a)*(x-c)+(x-b)*(x-c))=
K*((x-a)*(x-b)+(x-c)*(2x-a-b))
taking x=v=(a+b)/2 the second term vanishes and we have
f'(v)= K*(v-a)*(v-b)
the crossing point should be
v - f(v)/f'(v)
as in Newton's method and in this case should be:
v - {K*(v-a)*(v-b)*(v-c)}/{K*(v-a)*(v-b)}
= v - (v - c) = c
Deriving f(x)=K*(x-a)*(x-b)*(x-c) using the product rule we get
f'(x)= K*((x-a)*(x-b)+(x-a)*(x-c)+(x-b)*(x-c))=
K*((x-a)*(x-b)+(x-c)*(2x-a-b))
taking x=v=(a+b)/2 the second term vanishes and we have
f'(v)= K*(v-a)*(v-b)
the crossing point should be
v - f(v)/f'(v)
as in Newton's method and in this case should be:
v - {K*(v-a)*(v-b)*(v-c)}/{K*(v-a)*(v-b)}
= v - (v - c) = c
Comments on proofs of Euler's identity
You can only proof this identity after you clearly defined what e^(ix) means.
In this sense arguments using calculus and using differential equations are incomplete, without at least a clear definition any argument becomes an heuristic argument.
I know of three ways of defining e^(ix).
(1) From the complex series of e^(z) defined as sum of z^n/n!
(2) as the limit of n to infinity of (1 + ix/n)^n
(3) or directly as cos(x) + i sin(x)
Either way only after defining e^(ix) is that you should show that it has the properties like e^(ia)*e^(ib)=e^(i(a+b)), or (e^ix)'=ie^(ix) that you would expect it to have, and some use them fearlessly.
Number (1) is the most common one, I think number (2) would be a nice thing to use since it is analogous to the real case and can also be interpreted geometrically (as Richard Feynman does for a reference). But using (3) is totally misleading because it doesn't show why it should be true.
That is why I think heuristic arguments are needed to provide "a reason" for us to believe that such a thing should be true. Using circular motion as in my next post seems to me much more simple and much less "out of the blue". While the proof by itself is heuristic it can be made rigorous.
In this sense arguments using calculus and using differential equations are incomplete, without at least a clear definition any argument becomes an heuristic argument.
I know of three ways of defining e^(ix).
(1) From the complex series of e^(z) defined as sum of z^n/n!
(2) as the limit of n to infinity of (1 + ix/n)^n
(3) or directly as cos(x) + i sin(x)
Either way only after defining e^(ix) is that you should show that it has the properties like e^(ia)*e^(ib)=e^(i(a+b)), or (e^ix)'=ie^(ix) that you would expect it to have, and some use them fearlessly.
Number (1) is the most common one, I think number (2) would be a nice thing to use since it is analogous to the real case and can also be interpreted geometrically (as Richard Feynman does for a reference). But using (3) is totally misleading because it doesn't show why it should be true.
That is why I think heuristic arguments are needed to provide "a reason" for us to believe that such a thing should be true. Using circular motion as in my next post seems to me much more simple and much less "out of the blue". While the proof by itself is heuristic it can be made rigorous.
Proof of Euler's formula
This is a proof of Euler's formula that uses the circular motion
Suppose
P(t)= e^(it)
is a trajectory on the complex plane, using some definition of the exponential (the Taylor series definition for example) it is possible to show that
P'(t)=(e^(it))'= i e^(it) =i P(t)
and
P(0)= e^(i0)= e^(0)=1
In fact any reasonable definition of exponential should have these properties if you want to take these for granted all this becomes an heuristic proof.
then P(t) describes a trajectory in the complex plane that at t = 0 is at 1 and which velocity P(t) is always perpendicular, (90 degrees counter-clockwise) to P(t) and the velocity P(t) also has the same modulus as P(t). Such trajectory can only be the circular movement with radius one, unit velocity, starting at the point 1: cos(t) + i sin(t)!! If you want rigor again use the uniqueness of solution of differential equations the Picard–Lindelöf theorem.
e^(it)= cos(t) + i sin(t).
Suppose
P(t)= e^(it)
is a trajectory on the complex plane, using some definition of the exponential (the Taylor series definition for example) it is possible to show that
P'(t)=(e^(it))'= i e^(it) =i P(t)
and
P(0)= e^(i0)= e^(0)=1
In fact any reasonable definition of exponential should have these properties if you want to take these for granted all this becomes an heuristic proof.
then P(t) describes a trajectory in the complex plane that at t = 0 is at 1 and which velocity P(t) is always perpendicular, (90 degrees counter-clockwise) to P(t) and the velocity P(t) also has the same modulus as P(t). Such trajectory can only be the circular movement with radius one, unit velocity, starting at the point 1: cos(t) + i sin(t)!! If you want rigor again use the uniqueness of solution of differential equations the Picard–Lindelöf theorem.
e^(it)= cos(t) + i sin(t).
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