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Oct
13
comment How do I choose between $\lim_{x\to a} \frac {f(x) - f(a)}{x-a}\ $ and $\lim_{x\to a} \frac{f(a+h)-f(a)}{h}$?
To avoid the symbolic "overkill" you could always write the short hand notation for the binomial expansion, e.g. $$(x+h)^4=\sum_{k=0}^4{4\choose k}x^{4-k}h^k,$$ especially for large powers.
Oct
12
comment Find the limit of a Riemann Sum
$ f (X)=(1-X)(1+X) $.
Oct
9
accepted Are these functions identical?
Oct
8
asked Is it possible to abstract a Riemann integral into a “higher” integral with measure?
Oct
8
answered How to calculate $\sum\limits_{k=0}^{n}{k\dbinom{n}{k}}$
Oct
7
comment Calculating a complex definite improper integral: $I= \int_{0}^\infty x^{it}\,\mathrm{e}^{-ax}\, dx$
For a general treatment of these types of integrals, you may be interested in learning about the Bilateral Laplace Transform: en.wikipedia.org/wiki/…
Oct
3
comment how to calculuate $\int_0^ \pi \sqrt{1+x^2 \sin^2x}dx$
For what it's worth the integrand is symmetric, so we may write $$I = \frac{1}{2}\int_{-\pi}^\pi\sqrt{1+(x\sin x)^2}dx.$$
Oct
2
reviewed Approve suggested edit on Algebraic Integers and Irreducible Polynomials
Oct
2
comment Change of variable in complex integral
Your substitution should be $z=Re^{i\theta}$, so that $dz=Rie^{i\theta}d\theta$, and $0\leq \theta< \pi/4$. We make this substitution because the substitution describes the path $\Gamma$ which you give.
Oct
2
comment What is the relation between two integrals?
On the real line you would have $I_1\leq I_2$. But from your language it's unclear whether you're talking about integration over $\mathbb{R}$ or $\mathbb{C}$.
Oct
2
awarded  Nice Answer
Sep
30
awarded  Explainer
Sep
30
reviewed Approve suggested edit on How to proof the following function is always constant which satisfies $f\left( x \right) + a\int_{x - 1}^x {f\left( t \right)\,dt} $?
Sep
30
revised Why does being holomorphic imply so much about a function?
added 367 characters in body
Sep
27
accepted Can I use the residue calculus here?
Sep
27
comment Can I use the residue calculus here?
Thanks. I think you're right about the semi-circular contour. I also tried the box contour, and a similar problem arises. Maybe the residue calculus can't be used here.
Sep
27
comment Rationalise $\frac{2}{\sqrt{12}}$ fully
and $$\frac{\sqrt{12}}{6}=\frac{\sqrt{3\times 4}}{6}=\frac{2\sqrt{3}}{6}=\frac{\sqrt{3}}{3}.$$
Sep
27
accepted A question on the Wronskian
Sep
27
asked A question on the Wronskian
Sep
26
comment Why does being holomorphic imply so much about a function?
@Shakespeare The proof of the CR equations begins by thinking about all the possible directions. It can, however, be shown that the CR equations cover all such cases.