Integral and differential of summation

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SUMMARY

The discussion confirms two mathematical identities related to the integral and differential of summation. The first identity states that the derivative of a summation can be expressed as the summation of the derivatives, specifically: $$\frac{d}{dx} \sum_{u_0}^{u_1}f(x,u)\Delta u = \sum_{u_0}^{u_1}\frac{d}{dx}f(x,u)\Delta u$$. The second identity indicates that the integral of a summation equals the summation of the integrals: $$\int \sum_{u_0}^{u_1}f(x,u)\Delta u dx = \sum_{u_0}^{u_1}\int f(x,u)dx\Delta u$$. Both identities hold true under the condition that (u_1 - u_0)/Δx is finite, and their validity can be established through mathematical induction.

PREREQUISITES
  • Understanding of calculus, specifically differentiation and integration.
  • Familiarity with summation notation and its properties.
  • Knowledge of mathematical induction as a proof technique.
  • Basic comprehension of functions of multiple variables.
NEXT STEPS
  • Study the principles of mathematical induction in depth.
  • Explore advanced calculus topics, focusing on differentiation under the integral sign.
  • Learn about the properties of summation and their applications in calculus.
  • Investigate the implications of these identities in real-world scenarios, such as numerical methods.
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Mathematicians, students studying calculus, educators teaching advanced mathematics, and anyone interested in the theoretical foundations of calculus and summation techniques.

Jhenrique
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The following identities are true?
$$\frac{d}{dx} \sum_{u_0}^{u_1}f(x,u)\Delta u = \sum_{u_0}^{u_1}\frac{d}{dx}f(x,u)\Delta u$$ $$\int \sum_{u_0}^{u_1}f(x,u)\Delta u dx = \sum_{u_0}^{u_1}\int f(x,u)dx\Delta u$$
 
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As long as [itex](u_1- u_0)/\Delta x[/itex] is finite, those can be proved by induction.
 
HallsofIvy said:
As long as [itex](u_1- u_0)/\Delta x[/itex] is finite, those can be proved by induction.

I didn't understand...
 

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