bliengme

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MaplePrimes Activity


These are replies submitted by bliengme

@Preben Alsholm 

Changing Q[inf] to Qinf (no subscript) helped but I still cannot simplify to get an exp term
 

restart

"P(t):=(r*Qinf)/(2+2 cosh(b-r*t))"

proc (t) options operator, arrow; r*Qinf/(2+2*cosh(b-r*t)) end proc

(1)

Q := int(P(t), t = 0 .. t)

(1/2)*Qinf*tanh((1/2)*b)-(1/2)*Qinf*tanh(-(1/2)*r*t+(1/2)*b)

(2)

simplify(Q)

-(1/2)*Qinf*(-tanh((1/2)*b)+tanh(-(1/2)*r*t+(1/2)*b))

(3)

normal(Q)

(1/2)*Qinf*tanh((1/2)*b)-(1/2)*Qinf*tanh(-(1/2)*r*t+(1/2)*b)

(4)

``


 

Download Hubbert2.mw

 

@acer 

Many thanks; now I have better idea of what rand does

best wishes

@Kitonum many thanks for pointing out my foolish error!

@Kitonum 

I like this but I had tried that route - see my next post


 

@acer

Got it! I* really need to be more careful 

@acer 

This is excellent

@acer 

Like you I often think semilog means log scale on one axis, linear on the other

But others use the term for a scale that has logarithmic SPACING ; as in these two diagrams



@tomleslie Thanks but I want semilog not log. Have a look at Maple semilog to see what I mean but I wantit on the vertical xis

cheers

@acer Thanks for your quick reply - again!

I rely want semilog not log. Seems to be unavailable so I will use dualaxesplot

cheers

@acer 

I think I should be using the Maple zoom feature to help be see silly errors like this!

Cheers

 

@Joe Riel 

Outputs 2 and 7 are the same - good

Outputs 3 and 8 differ - bad but WHYTestBoth.mw
 

``

restart

u := -(1/2)*x^2-(1/6)*x^3+a*x+b; v := unapply(u, x)

-(1/2)*x^2-(1/6)*x^3+a*x+b

 

proc (x) options operator, arrow; -(1/2)*x^2-(1/6)*x^3+a*x+b end proc

(1)

s := solve({v(0) = 0, v(1) = 0}, {a, b})

{a = 2/3, b = 0}

(2)

assign(s)

a, b

2/3, 0

(3)

v(0); v(1)

0

 

0

(4)

restart

d2u := -1-x; d1u := int(d2u, x)+a; u := int(d1u, x)+b

-1-x

 

-x-(1/2)*x^2+a

 

-(1/2)*x^2-(1/6)*x^3+a*x+b

(5)

v := unapply(u, x)

proc (x) options operator, arrow; -(1/2)*x^2-(1/6)*x^3+a*x+b end proc

(6)

s = solve({v(0) = 0, v(1) = 0}, {a, b}); assign(s)

s = {a = 2/3, b = 0}

(7)

a, b

a, b

(8)

v(0); v(1)

b

 

-2/3+a+b

(9)

``


 

Download TestBoth.mwTestBoth.mw

 

@Kitonum 

Your reply requires me to replace the a and b by the values from solve. But in the top part of the worksheet that happened automatically. My question is why the lower part does not behave the same as the top part

@Kitonum 

I tried it and it worked

But Help does not show me the syntax for making a simplification and a change.

Can you point me to a place with info?

@Kitonum Did not know abot second paramenter in simplify - I must go back to Help

Thnak you

@acer Did not know about algsubs. Many thanks

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