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first summation vs consider the first summation

Both phrases are correct, but they are used in different contexts. 'First summation' is a concise way to refer to the initial calculation, while 'consider the first summation' is a more formal and detailed instruction to think about or analyze the initial calculation.

Last updated: March 15, 2024 • 494 views

first summation

This phrase is correct and commonly used to refer to the initial calculation in a series of sums.

This phrase is used to denote the initial calculation in a series of sums or mathematical operations.
  • ... of the second summation is simply factoring 3 from each term in the summation . Now apply Rule 1 to the first summation and Rule 2 to the second summation.).
  • 3, we then evaluate the entire expression governed by the first summation op- erator. This is the expression contained in large parentheses. This expression.
  • First, summation notation. In summation notation, you are given an expression and told how many terms you are to add up. tells you to use the values of n = 1, ...
  • Calculation. Previous day's Summation Index* + current day McClellan Oscillator * The very first Summation Index is simply the value of the McClellan Oscillator.

consider the first summation

This phrase is correct and can be used when instructing someone to think about or analyze the initial calculation in a series of sums.

This phrase is used in a more formal context to instruct someone to think about or analyze the initial calculation in a series of sums or mathematical operations.
  • The first step is to re-index the powers of x as they appear in each summation so that all powers of x are the same. Let's consider the first summation on the left: ...
  • Oct 15, 2010 ... ... has a nice proba- bilistic interpretation which allows us to obtain a closed-form formula. To be more precise, consider the first summation. ∑.
  • +dV (µ2 ∗ ν1,ν2 ∗ ν1). = ∑ z∈Z. |µ1 ∗ µ2(z) − ν1 ∗ µ2(z)|. +. ∑ z∈Z. |µ2 ∗ ν1(z) − ν2 ∗ ν1(z)|. Consider the first summation. ∑ z∈Z. |µ1 ∗ µ2(z) − ν1 ∗ µ2(z)|. =.
  • x(n)eLj2πn(2*M/M3)/N. Now consider the first summation and rearrange the terms in pairs (separated by N'/2 terms): N3L1. Σ n=0 x(n)eLj2πn(2*M/M3)/N = (6) . =.

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