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andreev551 [17]
3 years ago
13

Which property is shown in the following statement? 7 · 1 = 7 A. identity property of multiplication B. associative property of

multiplication C. commutative property of multiplication
Mathematics
2 answers:
Elan Coil [88]3 years ago
7 0

Answer:

identity property of multiplication

Step-by-step explanation:

Identity property of multiplication means the identity stays the same. in other words to multiply by 1.

dangina [55]3 years ago
7 0
This is an example of the identity property of multiplication. The basic meaning of it is that any number multiplied by one is equal to itself. Associative means you group numbers together when multiplying and commutative is multiplying in any order
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Find the domain of the Bessel function of order 0 defined by [infinity]J0(x) = Σ (−1)^nx^2n/ 2^2n(n!)^2 n = 0
Snowcat [4.5K]

Answer:

Following are the given series for all x:

Step-by-step explanation:

Given equation:

\bold{J_0(x)=\sum_{n=0}^{\infty}\frac{((-1)^{n}(x^{2n}))}{(2^{2n})(n!)^2}}\\

Let   the value a so, the value of a_n  and the value of a_(n+1)is:

\to  a_n=\frac{(-1)^2n x^{2n}}{2^{2n}(n!)^2}

\to a_{(n+1)}=\frac{(-1)^{n+1} x^{2(n+1)}}{2^{2(n+1)}((n+1))!^2}

To calculates its series we divide the above value:

\left | \frac{a_(n+1)}{a_n}\right |= \frac{\frac{(-1)^{n+1} x^{2(n+1)}}{2^{2(n+1)}((n+1))!^2}}{\frac{(-1)^2n x^{2n}}{2^{2n}(n!)^2}}\\\\

           = \left | \frac{(-1)^{n+1} x^{2(n+1)}}{2^{2(n+1)}((n+1))!^2} \cdot \frac {2^{2n}(n!)^2}{(-1)^2n x^{2n}} \right |

           = \left | \frac{ x^{2n+2}}{2^{2n+2}(n+1)!^2} \cdot \frac {2^{2n}(n!)^2}{x^{2n}} \right |

           = \left | \frac{ x^{2n+2}}{2^{2n+2}(n+1)^2 (n!)^2} \cdot \frac {2^{2n}(n!)^2}{x^{2n}} \right |\\\\= \left | \frac{x^{2n}\cdot x^2}{2^{2n} \cdot 2^2(n+1)^2 (n!)^2} \cdot \frac {2^{2n}(n!)^2}{x^{2n}} \right |\\\\

           = \frac{x^2}{2^2(n+1)^2}\longrightarrow 0   for all x

The final value of the converges series for all x.

8 0
3 years ago
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