Carl Love

Carl Love

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12 years, 106 days
Himself
Wayland, Massachusetts, United States
My name was formerly Carl Devore.

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These are replies submitted by Carl Love

Are the x_n restricted to positive reals?

Combining Axel's idea with my power series idea given in another Answer, we see that the power series expressed with lower limit n is also independent of n, and is the same as if the lower limit were 0. However, oddly enough, though the two expressions are equal, they are not identical as expressions.

S1:= sum(x^k/(6*k+1)!, k= 0..infinity);

S2:= sum(x^k/(6*k+1)!, k= n..infinity);

simplify(S1-S2);
                        0      

Combining Axel's idea with my power series idea given in another Answer, we see that the power series expressed with lower limit n is also independent of n, and is the same as if the lower limit were 0. However, oddly enough, though the two expressions are equal, they are not identical as expressions.

S1:= sum(x^k/(6*k+1)!, k= 0..infinity);

S2:= sum(x^k/(6*k+1)!, k= n..infinity);

simplify(S1-S2);
                        0      

But it seems that the series starting at k = 0 is viewed as a power series evaluated at 1:

sum(x^k/(6*k+1)!, k= 0..infinity):
eval(%, x=1);

But it seems that the series starting at k = 0 is viewed as a power series evaluated at 1:

sum(x^k/(6*k+1)!, k= 0..infinity):
eval(%, x=1);

It is not clear what you want.

  • Let A be the region enclosed by r = 1.
  • Let B be the region enclosed by r = 2*cos(theta)

Then, proceeding left to right, there are three disjoint regions:

  • Let R1 = A minus B
  • Let R2 = A intersect B
  • Let R3 = B minus A

What color do you want R1? What color do you want R2? What color do you want R3? What color do you want the boundary of A? What color do you want the boundary of B? Do you want to see the gridlines underneath the regions?

 

 

@litun Consider a simpler example: z(t) = t, G = 1 + t + t^2. Then G = z^0 + z^1 + z^2, so a_0(t) = a_1(t) = a_2(t) = 1. Or G = (1+t)*z^0 + t*z^1, so a_0(t) = 1 + t, a_1(t) = t, a_2(t) = 0. And there are numerous other possibilities. So the a_k(t) are not uniquely determined.

@tuGUTS Like I said, I don't know of any case where rsolve solves an equation with more than one independent variable; although I'm not absolutely sure that it will never solve them. Hopefully someone who knows more will comment on this.

This is just a seat-of-the-pants guess, but the answer to your equation may involve Stirling numbers.

Adri wrote:

...it is automatically filled with zeros on the empty places.

Zero is the default, but the fill value can be anything, including NULL.

b:= Matrix([a||(1..4)], fill= NULL);

The advantage of NULL is that it is automatically removed when converted to a list.

convert(b^%T, listlist);
 
[[1, 5, 4, 2], [2, 3, 8, 2], [4, 4, 5], [5, 5, 4], [3, 7], [2, 8], [3], [4], [5]]

Adri wrote:

...it is automatically filled with zeros on the empty places.

Zero is the default, but the fill value can be anything, including NULL.

b:= Matrix([a||(1..4)], fill= NULL);

The advantage of NULL is that it is automatically removed when converted to a list.

convert(b^%T, listlist);
 
[[1, 5, 4, 2], [2, 3, 8, 2], [4, 4, 5], [5, 5, 4], [3, 7], [2, 8], [3], [4], [5]]

@Christopher2222 wrote:

On interpretations of the same sentence.  
He didn't marry her because she was rich - (he didn't marry her)
He didn't marry her because she was rich - (he did marry her)

When the subordinate clause ("because she was rich") does not change the meaning of the main clause ("he didn't marry her"), it needs to be set off with a comma.

He didn't marry her, because she was rich. (He didn't marry her.)
He didn't marry her because she was rich. (He did marry her.)

Can you give a more specific example? I'm especially having trouble with the part about z being a function of t. And why is the answer not simply a_k(t)?

Both evalc and expand work for me in Maple 16.02 (and Maple 17). What does expand give you? The command made specifically for this purpose is evalc. The simplified value that you gave is wrong however.

Your last equation is not linear, assuming the Rs are the unknowns, because it has two Rs per term.

Please upload your worksheet, not an image. It is more than inconvenient for us to download the image: We can't be sure what you actually typed into Maple without having the worksheet.

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