D'oh, never mind, sorry! I missed the separate 10.10 .dmg near the bottom
of the download page.
On Friday, February 27, 2015 at 9:17:22 PM UTC-5, Murray Eisenberg wrote:
>
> I downloaded, verified MD5 checksum for, and
> opened sage-6.5-x86_64-Darwin-OSX_10.7_x86_64-app.dmg on my iMac running OS
I downloaded, verified MD5 checksum for, and
opened sage-6.5-x86_64-Darwin-OSX_10.7_x86_64-app.dmg on my iMac running OS
X 10.10.2, then moved the app into my Applications folder. After the usual
nod to Apple's security about the app, I opened it.
While Activity Monitor shows it's running, the
On 2015-02-27, j wade wrote:
> If I just enter
>
> solve(90*e^(-1)/c == 1.50, c)
>
> I again get [].
>
> But, if I enter
>
> b=var('b')
> solve(90*e^(-1)/b == 1.50, b)
>
> I get the solution:
>
> [b == 60*e^(-1)]
For the record, when I try this directly in Max
Thank you!
On Friday, February 27, 2015 at 5:41:07 PM UTC+2, vdelecroix wrote:
>
> Here is one way... not sure it is the best
>
> sage: eq1 = sqrt(cos(4*x)+1)
> sage: eq2 = eq1.simplify_trig()
> sage: eq2
> sqrt(8*cos(x)^4 - 8*cos(x)^2 + 2)
>
> The next step consists in factoring what is insi
Hello Everybody,
Sorry for posting the same question. I searched a lot but I didn't find
anything understandable. It might be because of my weakness in related
concepts.
I am trying to use Pycharm as an IDE for Sage codes. I read this
discussion, but It was not enough clear for me.
https://gr
I am having an odd issue with solve in Sage. I am using Sage 6.3 in Ubuntu
14.04.
Entering the following commands:
forget()
var('m k c x t')
m=3
assume(4*k*m-c^2==0)
x=function('x',t)
de = m*diff(x,t,2)+c*diff(x,t)+k*x==0
x=desolve(de,x,ivar=t)
x=x.subs(_K1=0)
x=x.subs(_K2=15)
solve(x(t=(6
Here is one way... not sure it is the best
sage: eq1 = sqrt(cos(4*x)+1)
sage: eq2 = eq1.simplify_trig()
sage: eq2
sqrt(8*cos(x)^4 - 8*cos(x)^2 + 2)
The next step consists in factoring what is inside the sqrt:
sage: o = eq2.operands()[0]
sage: of = o.factor()
sage: o
8*cos(x)^4 - 8*cos(x)^2 + 2
s
OK. Let x is real.
How to rewrite sqrt(cos(4x)+1) into sqrt(2)abs(cos(2x))?
On Friday, February 27, 2015 at 3:36:59 PM UTC+2, Simon King wrote:
>
> Hi Paul,
>
> On 2015-02-27, Paul Royik > wrote:
> > What is the way to consistently simplify square roots of squares?
> >
> > Examples:
> >
> >
But...
sage: eq = sqrt((pi-5)^2)
sage: eq.canonicalize_radical()
pi - 5
And as you can read from the documentation
"""
Choose a canonical branch of the given expression. The square root,
cube root, natural log, etc. functions are multi-valued. The
"canonicalize_radical()" method will cho
Hi Paul,
On 2015-02-27, Paul Royik wrote:
> What is the way to consistently simplify square roots of squares?
>
> Examples:
>
> sqrt((x+1)^2) - > x+1
> sqrt(cos(4*x)+1) -> sqrt(2)cos(2x)
Simplification must not change the value of the expression. sqrt(x^2) is
certainly not equal to x. Even under
What is the way to consistently simplify square roots of squares?
Examples:
sqrt((x+1)^2) - > x+1
sqrt(cos(4*x)+1) -> sqrt(2)cos(2x)
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