# Math ASCII Notation Demo

Mathematical content on Apronus.com is presented in Math ASCII Notation which can be properly displayed by all Web browsers because it uses only the basic set of characters found on all keyboards and in all fonts.

The purpose of these pages is to demonstrate the power of the Math ASCII Notation. In principle, it can be used to write mathematical content of any complexity. In practice, its limits can be seen when trying to write complicated formulas (containing, for example, variables with many indexes or multiple integrals).

Despite its limitations the Math ASCII Notation has much expressive power, as can be seen from browsing through these pages.

Where does the series log( 1 + x*x / n*log(n)*log(n) ) converge ?
(x is real)
converges for all x
hint:
log(1+x)<=x
2^k*a(2^k)
Is it true that when a power series with center 0 converges for z0, then it converges absolutely for |z|<|z0| ?
Yes. This proof is very easy.
Suppose that the limit |a[n]| / |a[n+1]| exists. Show that it is the radius of convergence of the power series a[n]*z^n.
Use the ratio test for convergence of series.
Prove that 1 / lim_sup |a[n]| ^ (1/n) is the radius of convergence of the power series a[n]*z^n.
Use the Cauchy test for convergence of series.
Find the radius of convergence of the series
x^n / ( 2^n + 3^n + (-1)^n * (3^n - 2^n) ).
R = 2
Use Cauchy test.
Find the radius of convergence of the series ( n! / 2^(n*n) ) * x^n.
R = oo.
the ratio test
Let p[n] be the sequence of consecutive prime numbers. Find the radius of convergence of the power series x^p[n] / p[n].
R = 1
Notice that the disputed series converges for -1.
Hence the radius must be at least 1.
Notice that if a>1, then the series diverges for a.
Hence the radius cannot be greater than 1.
Conclusion: R=1
Let p[n] be the sequence of consecutive prime numbers. Find the radius of convergence of the power series p[n] * x^p[n].
R = 1
way1:
a[n] = { n for prime n; 0 otherwise }
Use the Cauchy test.
way2:
0<x<1 => +(n=1 to oo) p[n]*x^p[n] < +(n=1 to oo) n*x^n < oo
Where does this series converge?
1/(x^n) * sin(1/(2^n))
|x| > 1/2
Hint: lim sin(x)/x = 1 as x approaches 0.
By referring to the Euclidean two-dimensional space, justify intuitively why
[A x B = 0] <=> [A,B are perpendicular].
A x B is the scalar product (a,b)x(c,d) = (a*c,b*d)
Hint: |A-B| = |A+B|