Owen's Math Portal

For a set of orthonormal vectors \((v_1,v_2,...,v_m)\) in an m-dimensional vector space , and set of constants \(r_1,r_2,...r_m \in \mathbb{R}\) , if we consider the equation \(r_1v_1+r_2v_2+...+r_mv_m=0\) with the objective of demonstrating that the vectors in \((v_1,v_2,...,v_m)\) are linearly independent, then we must show that \(r_1=r_2=...=r_m=0\) . If we take the inner product on both sides of the equation , we have that $$ <r_1v_1+r_2v_2+…+r_mv_m, v_1> = <0,v_1> $$ Using the properties of inner products, we have that

We will evalute the integral \(\int_0^1 \frac{\ln(x)}{1+x} dx \) using Taylor Series expansion and switching the order of integration and summation. Firstly, we notice that $$ \frac{1}{1+x}=\sum_{n=0}^{\infty} (-1)^n x^n $$ by taylor series expansion. This series coverges for \(x \in (0,1)\) so we can substitute this into the integral: $$ \int_0^1 \frac{\ln(x)}{1+x} dx = \int_0^1 \ln(x) \sum_{n=0}^{\infty} (-1)^n x^n dx $$ Further, by changing the order of integration and summation, we have $$ \int_0^1 \ln(x) \sum_{n=0}^{\infty} (-1)^n x^n dx = \sum_{n=0}^{\infty} (-1)^n \int_0^1 x^n \ln(x) dx $$

The Basel Problem, named after Leonard Euler’s hometown of Basel, Switzerland, is the solution to the infinite series \(\sum_{n=1}^{\infty} \frac{\pi^2}{6}\) Euler was the first to publish his solution one year after having solved this problem in 1734. Since Euler’s solution, many alternative proofs have been offered, using complex analysis, symmetric polynomials, and so on. These alternative proofs are more rigorous than Euler’s liberal use of infinite polyomial tails, but Euler’s solution is perhaps the most elegant.