## 89 Reputation

16 years, 147 days

## Much appreciated, acer!...

Much appreciated, acer!

## Much appreciated, acer!...

Much appreciated, acer!

## Greatly appreciate your...

Greatly appreciate your assistance. To conclude, I await your comment on the explanation for XOR algorithm I provided above in my response. Look forward. Best regards, wirefree

## Greatly appreciate your...

Greatly appreciate your assistance. To conclude, I await your comment on the explanation for XOR algorithm I provided above in my response. Look forward. Best regards, wirefree

## "Is it more clear now?"...

"Is it more clear now?" A lot more, Alec. Appreciate that. However, my understanding is only as deep as my ability to arrive at the 2^22 figure for the xOR alternative you have suggested. I attempt to provide a framework below and would appreciate if you could comment on its accuracy & completeness. a) 2^12 is arrived at in a similar fashion, by considering all possibilities for the highest 6 bits. b) 2 choices for 1 (1 xor 0, 0 xor 1) and 2 choices for 0 (0 xor 0, 1 xor 1) yields 2^10, since there exist 2 choices for every position. c) This yields a total of 2^22 IP addresses mapped to every number from 0 to 1023. Please comment on the above rationale. With regards to formal requirements of the question, namely, plot of the CDF and giving 1st and 2nd moments - would appreciate your suggestion on how may these be arrived at in Maple vis-a-vis the suggested framework. Look forward to your response. Best regards, wirefree

## "Is it more clear now?"...

"Is it more clear now?" A lot more, Alec. Appreciate that. However, my understanding is only as deep as my ability to arrive at the 2^22 figure for the xOR alternative you have suggested. I attempt to provide a framework below and would appreciate if you could comment on its accuracy & completeness. a) 2^12 is arrived at in a similar fashion, by considering all possibilities for the highest 6 bits. b) 2 choices for 1 (1 xor 0, 0 xor 1) and 2 choices for 0 (0 xor 0, 1 xor 1) yields 2^10, since there exist 2 choices for every position. c) This yields a total of 2^22 IP addresses mapped to every number from 0 to 1023. Please comment on the above rationale. With regards to formal requirements of the question, namely, plot of the CDF and giving 1st and 2nd moments - would appreciate your suggestion on how may these be arrived at in Maple vis-a-vis the suggested framework. Look forward to your response. Best regards, wirefree

Appreciate your detailed response, Alec. The simplicity of your formulation is very appealing. I, however, cannot avail of it due to shortcomings in my understanding. If I may be allowed to quote from your response and decompose the solution so as to render it addressable, I would request you to fill in blanks. "A 16-bit number mod 1024 = 2^10 is just the lowest 10 bit of it." -> My understaning of mod is only restricted to remainder of a division operation. I need to understand about bit-wise mod. What yields the lowest 10 bits only? "The highest 6 bits are not used, so the 12 bits that they are made from, can be anything." -> Were you referring to the '10' bits from the quote above? Which 12 bits are these? Since every 0 bit can be obtained in 3 cases, and 1 - in 1 case, by the product rule, the number of IP addresses mapping to n equals 3^z(n)*2^12. -> Although the explanation here may be sufficient, some labor through the details would prove beneficial. Look forward to your promot response. Best regards, wirefree

Appreciate your detailed response, Alec. The simplicity of your formulation is very appealing. I, however, cannot avail of it due to shortcomings in my understanding. If I may be allowed to quote from your response and decompose the solution so as to render it addressable, I would request you to fill in blanks. "A 16-bit number mod 1024 = 2^10 is just the lowest 10 bit of it." -> My understaning of mod is only restricted to remainder of a division operation. I need to understand about bit-wise mod. What yields the lowest 10 bits only? "The highest 6 bits are not used, so the 12 bits that they are made from, can be anything." -> Were you referring to the '10' bits from the quote above? Which 12 bits are these? Since every 0 bit can be obtained in 3 cases, and 1 - in 1 case, by the product rule, the number of IP addresses mapping to n equals 3^z(n)*2^12. -> Although the explanation here may be sufficient, some labor through the details would prove beneficial. Look forward to your promot response. Best regards, wirefree

## Hello there, How may I plot...

Hello there, How may I plot distributions such as the following: with(Statistics): X := RandomVariable(Beta(nu,omega)): PDF(X, u); Would appreciate advise for Maple11. Best regards, wirefree

## Appreciate the response. I...

Appreciate the response.

I am clear on the order query. However, when I refer to sigma, I mean the following type format of series solutions:

Professor has suggested exploring "formal series" parameter in dsolve. But, maple is unable to process the command. The commands I issue are:

ode := (x^2+2)*(diff(diff(y(x), x), x))+3*x*(diff(y(x), x))-y(x) = 0;

dsolve(ode, y(x), 'formal_series', 'coeffs' = 'polynomial');

Would appreciate feedback.

Best,

wirefree

## Appreciate the response. I...

Appreciate the response.

I am clear on the order query. However, when I refer to sigma, I mean the following type format of series solutions:

Professor has suggested exploring "formal series" parameter in dsolve. But, maple is unable to process the command. The commands I issue are:

ode := (x^2+2)*(diff(diff(y(x), x), x))+3*x*(diff(y(x), x))-y(x) = 0;

dsolve(ode, y(x), 'formal_series', 'coeffs' = 'polynomial');

Would appreciate feedback.

Best,

wirefree

## Hi,...

Hi,

Is there a procedure to obtain results from dsolve in the sigma format?

Alternatively, how can Maple present more terms? For instance, the case of dsolve on ((D@@2)(y))(x)-x*y(x) = 0 , results in more than just 2 terms each for c0 & c1.

Would appreciate a response.

Best,

wirefree

## Hi,...

Hi,

Is there a procedure to obtain results from dsolve in the sigma format?

Alternatively, how can Maple present more terms? For instance, the case of dsolve on ((D@@2)(y))(x)-x*y(x) = 0 , results in more than just 2 terms each for c0 & c1.

Would appreciate a response.

Best,

wirefree

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