- Transforming a sum to an integral: why does it work?
By assuming the separation distance between the points in the discrete space is negligibly small compared to the total volume, we can make use of the definition of the Riemann integral Say we are summing over discrete points k in k-space ∑ k f(k)
- Replace a sum with an integral $\\sum\\rightarrow \\int$
How can one turn a sum to an integral Example $$\sum_k f(k) \approx N\cdot\int_k dk\, f(k) $$ How do you find the factor $N$? The quantities should be approximately equal
- Can you simply replace an integral with a summation?
Can you really just swap a sum and an integral? I mean, the sum of a linear variable k is a second degree polynomail, and the integral of a continuous variable k would be a 2nd degree polynomial, so maybe there's something to it The answer is yes, under certain circumstances
- 13. 3 Tricks of Integration - MIT Mathematics
The chain rule tells us how to differentiate \(f(g(x))\) and the answer is \(\frac{df}{dg}\frac{dg}{dx}\) This tells us that we if we can recognize an integrand as having the form \(\frac{df}{dg}g'\), we can integrate it over \(dx\) to get \(f(g(x))\) evaluated at \(b\) less its evaluation at \(a\) What can we recognize this way?
- Notes on the Fundamental Theorem of Integral Calculus
Then the integral over the entire circle γ is the sum of the integrals over these semicircles − C\{(−∞, 0]} (make a sketch of G so that you can see its relationship with γ1) Then F is analytic in G and γ1 lies entirely in G, so by our Fundamental Theorem for Line Integrals:
- MATH 409 Advanced Calculus I Lecture 33: Propertiesofthe . . .
Proof: Since f ≤ g on the interval [a,b], it follows that S(f,P,t j) ≤ S(g,P,t j) for any partition P of [a,b] and choice of samples t j As kPk → 0, the sum S(f,P,t j) gets arbitrarily close to the integral of f while S(g,P,t j) gets arbitrarily close to the integral of g The theorem follows Corollary If f is integrable on [a,b] and f
- integration - Requirement for turning sum into integral . . .
The Riemann sum $\Delta k\sum_k f(k)$ converges to $\int f(k)\text dk$ in the limit $\Delta k\to 0$ To prove this, consider a small cell $D_k$ of length $\Delta k$ around each value of $k$ in the sum
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