taro

445 Reputation

12 Badges

6 years, 233 days
Maple is to me difficult. The first version I bought was Maple9, and it was more than 15 years ago. But, I couldn't use it, feeling it too difficult. But, three years ago, I thought Maple might be helpful to my study, and since then, I have continued to learn Maple. As I got able to read the Maple help, I think that I could get to use maple better now than before. But, I feel that I am a beginner yet.

MaplePrimes Activity


These are replies submitted by taro

@Carl Love 

Thank you alway for your teaching.

I owe you quite a lot.

And, though I'm sorry for asking for futher answer,

I have another questions about your answer.

in your code of

proc():-`print/h`:= ()-> ':-h'(args)

why ":-" comes after proc()? This is the first time I see the form, with proc taking

statement after it. And, why there occurrs no problem even if including ':-' in front of h.

I can't understand the meaning of the form of ':-h'.

Definitly the day would not come, when I hit on your answer, for myself.

Thank you for your helping me a lot.

Best wishes.

taro

I brought maple back to 10.11 of OSX.

I had been worrying about mail. But, though that operation as well, there wasn't the case that

mail had dissapeared. Of course, before starting that operation, I backed up the contents of this macbook

to a hard disk.

 

 

 

@acer 

I had been used to updates of software when OS got new.

Surely, it is strange as updating software needs cost and time.

Thank you. I will wait for some update if there would be, or for the next version of maple.

Best wishes.

taro 

@vv 

Surely, if this were 2D, there does not look having a gap.

z1, z2 := solve(x^2+y^2 = 1, {x, y})

plot([rhs(z1[1]), rhs(z2[1])], y = -1 .. 1, scaling = constrained, color = red);

 

As for the range of your plot, you can determine it with

for example

solve(eval((x-y)^2+(x-z)^2+(z-x)^2 = 3, {x = 2, y = 2}), z):evalf(%);

 

z1 and z2 have z,k1 and k2 as unknowns, and uu has alpha, z1 and z2, so that alpha, k1 and k2 as unknowns.

You are talking about intersections. But, between which intersection are you talking of?

 

 

 

@Christian Wolinski 

Is your link to

http://www.mapleprimes.com/questions/210560-Output-List-From-Op-Command

?

Thank you for the above link. I will try to understand it.

Best wishes.

taro

@Preben Alsholm 

Thank you for teaching me about expanding a denominator.

I could know owing to you that normal with expanded is the specific command for it.

And, with applyop, I could know that it is possible to limit the part expanded to a part of an expression.

Thank you.

Best wishes.

taro

 

@Carl Love 

Thank you very much for your always teaching me about Maple.

I could know how to use evalindets owing to you. And, I couldn't believe that its use with d-> op(d)*'h'(op(d))

 will bring the result which I want to obtain. It looks a magic to me.

h(x), which means dx/x, is the expression frequently used in the field I study in economics.

So, I had wondered whether Maple can do that calculation. And, I had thought that

if it was impossible, there might have been the limit of using CAS including Maple in reading and writing papers.

But, as I had been still a newbie in using maple and am so still now, I had thought that the time I would find how to

do that logarithmic differentiation would be in the unknownablely further future, even if there would be.

Thank you for teaching how to do logarithmic differentiation with maple to me.

 

Best wishes.

 

taro

 

 

 

 

@Markiyan Hirnyk 

 h(x) which I want to have is  a short form D(x)/x, where D(x) is a total derivative of x, that is dx.

And, i want to modify about x+y  with "h" as

h(x+y)=(x/(x+y))*h(x)+(y/(x+y))*h(y).

And, I want to use that function "h" as h(f(x)*y+x*g(y)) as an input. Then, that function h

returns { (f(x)*y)/(f(x)*y+x*g(y))}*h(f(x)*y) + {x*g(y)/(f(x)*y+x*g(y))}*h(x*g(y))), and

further, it is modified to { (f(x)*y)/(f(x)*y+x*g(y))}*(h(f(x))+h(y))+{x*g(y)/(f(x)*y+x*g(y))}*{h(x)+h(g(y))}.

The expression like D(log(x)) brabra is somewhat long when it is shown on the screen.

Could I make it clear to you what I should make clear? I hope I did so.

@Carl Love

As h(x) is intended as D(x)/x, where D(x) means dx, that is a total derivative of x,

so h(x+y)=(x/(x+y))*h(x)+(y/(x+y))*h(y) means

(x/(x+y))*D(x)/x + (y/(x+y))*D(y)/y, which is D(x)/(x+y) + D(y)/(x+y).

I wrote D(x) for dx, total derivative of x.

And, d(x+y) = dx + dy. So, d(x+y)/(x+y), which is h(x+y), becomes dx/(x+y) + dy/(x+y), which is

what I wrote D(x)/(x+y) + D(y)/(x+y).

 

Best wishes.

taro

@Preben Alsholm 

Thank you very much.

I could understand what you explain.

 

Best wishes.

taro

 

 

 

 

 

 

@Carl Love 

It's because almost everyday I report spam, but as I cannot delete them, they remain on the board even after my

report. If a lot more people can delete spams, opportunities a lot of people are worried by the same spams becomes

less.

taro

 

As retrieving deleted posts is said to be easy, I hope that giving a measure to delete posts, not reporting as a spam,

to people with higher than 100 will be considered.

taro

@Kitonum

Thank you for your answer.

@Carl Love 

Thank you. `n.d` is a symbol in itself.

So, though `convert/string`(n.d) surely returns "n . d", it is natural that `convert/string`(`n.d`) returns "`n.d`".

But, this does not explain why convert(`n.d`,string); retuns "n.d". There must be another reason for it.

But, I will finish my question with my writing of this.

Best wishes.

taro

 

 

 

 

@Kitonum 

Please teach me why

`convert/string`(`N.D`);

and

`convert/string`(`Norman`);

bring different results, with the former  "`N.D`" while the latter "Norman" without a back quotation.

 

 

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