Why substitution works in indefinite integration

Let’s integrate . We know the trick: substitute for . We get . Substituting into the original equation, we get . Let us assume remains positive throughout the interval under consideration. Then we get the integral as or . I have performed similar operations for close to five years of my life now. But IContinue reading “Why substitution works in indefinite integration”

Fermat’s Last Theorem

When in high school, spurred by Mr. Scheelbeek’s end-of-term inspirational lecture on Fermat’s Last Theorem, I tried proving the same for…about one and a half long years! For documentation purposes, I’m attaching my proof. Feel free to outline the flaws in the comments section. Let us assume FLT is true. i.e. . We know (Continue reading “Fermat’s Last Theorem”

Binomial probability distribution

What exactly is binomial distribution? Q. A manufacturing process is estimated to produce nonconforming items. If a random sample of the five items is chosen, find the probability of getting two nonconforming items. Now one could say let there be items. Then the required probability woud be . In what order the items are chosenContinue reading “Binomial probability distribution”

Continuous linear operators are bounded.: decoding the proof, and how the mathematician chances upon it

Here we try to prove that a linear operator, if continuous, is bounded. Continuity implies: for any for We want the following result: , where is a constant, and is any vector in . What constants can be construed from and , knowing that they are prone to change? As is a linear operator, isContinue reading “Continuous linear operators are bounded.: decoding the proof, and how the mathematician chances upon it”

Linear operators mapping finite dimensional vector spaces are bounded,

Theorem: Every linear operator , where is finite dimensional, is bounded. Proof where . What we learn from here is where . Similarly, where Another proof of the assertion is which is a constant. Note: why does this not work in infinite dimensional spaces? Because the max and min of and might not exist.

Completing metric spaces

If you’ve read the proof of the “completion of a metric space”, then you surely must have asked yourself “WHY?”! Say we have an incomplete metric space . Why can’t we just complete by including the limit points of all its cauchy sequences?! No. We can’t. The limit points of cauchy sequences may not beContinue reading “Completing metric spaces”

A new proof of Cauchy’s theorem

We will discuss a more direct proof of Cauchy’s theorem than the one given in Herstein’s “Topics in Algebra” (pg.61). Statement: If is an abelian group, and , then there is an element such that , and . We will prove this by induction. Let us assume that in every abelian group of order ,Continue reading “A new proof of Cauchy’s theorem”

Today we will discuss compactness in the metric setting. Why metric? Because metric spaces lend themselves more easily to visualisation than other spaces. Let us imagine a metric space with points scattered all over it. If we can find an infinite number of such points and construct disjoint open sets centred on them, then cannotContinue reading