How to use AND condition in Desmos
Sorry maybe it's not typical mathematics question, but Desmos is very helpful in solving and testing mathematics issues, so maybe anyone could help me.
I can't figure it out how to use AND condition in Desmos
For example to make OR you can just use "comma", like that:
$ y={ A<0 , B>1 : f_1(x), f_2(x) } $
But how to make AND? I mean something like that:
$ y={ A<0 AND B>1 : f_1(x), f_2(x) } $
But that just doesn't work.
For any help thanks in advance.
graphing-functions conditional-convergence
asked Nov 26 at 13:34
pajczur
854
add a comment |
Sorry maybe it's not typical mathematics question, but Desmos is very helpful in solving and testing mathematics issues, so maybe anyone could help me.
I can't figure it out how to use AND condition in Desmos
For example to make OR you can just use "comma", like that:
$ y={ A<0 , B>1 : f_1(x), f_2(x) } $
But how to make AND? I mean something like that:
$ y={ A<0 AND B>1 : f_1(x), f_2(x) } $
But that just doesn't work.
For any help thanks in advance.
graphing-functions conditional-convergence
asked Nov 26 at 13:34
pajczur
854
add a comment |
Sorry maybe it's not typical mathematics question, but Desmos is very helpful in solving and testing mathematics issues, so maybe anyone could help me.
I can't figure it out how to use AND condition in Desmos
For example to make OR you can just use "comma", like that:
$ y={ A<0 , B>1 : f_1(x), f_2(x) } $
But how to make AND? I mean something like that:
$ y={ A<0 AND B>1 : f_1(x), f_2(x) } $
But that just doesn't work.
For any help thanks in advance.
graphing-functions conditional-convergence
asked Nov 26 at 13:34
pajczur
854
Sorry maybe it's not typical mathematics question, but Desmos is very helpful in solving and testing mathematics issues, so maybe anyone could help me.
I can't figure it out how to use AND condition in Desmos
For example to make OR you can just use "comma", like that:
$ y={ A<0 , B>1 : f_1(x), f_2(x) } $
But how to make AND? I mean something like that:
$ y={ A<0 AND B>1 : f_1(x), f_2(x) } $
But that just doesn't work.
For any help thanks in advance.
graphing-functions conditional-convergence
graphing-functions conditional-convergence
asked Nov 26 at 13:34
pajczur
854
asked Nov 26 at 13:34
pajczur
854
asked Nov 26 at 13:34
pajczur
854
asked Nov 26 at 13:34
pajczur
854
asked Nov 26 at 13:34
pajczur
854
854
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
I haven't found a built-in conjunction function, so you might find helpful my workaround:
For simple domain restrictions, you can use a product of conditions:
$$y=f(x){text{cond1}}{text{cond2}}{text{cond3}}$$
$$y=(x^2-x)left{x^2>1right}left{x^4<3right}$$
link
For complex piecewise functions, you can use this hack:
$$text{cond1} = left({text{subcond1}:1,0}{text{subcond2:1,0}}{text{subcond3:1,0}}=1right)$$
$$
y=left{\color{red}{left{x^2>1:1,0right}left{x^4<3:1,0right}=1:left(x^2-xright)},\color{blue}{left{x^2>1.8:1,0right}left{x^4+x^2<25:1,0right}=1:x},\color{green}{-1}right}$$
link
answered Nov 26 at 17:20
Vasily Mitch
1,32837
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
I haven't found a built-in conjunction function, so you might find helpful my workaround:
For simple domain restrictions, you can use a product of conditions:
$$y=f(x){text{cond1}}{text{cond2}}{text{cond3}}$$
$$y=(x^2-x)left{x^2>1right}left{x^4<3right}$$
link
For complex piecewise functions, you can use this hack:
$$text{cond1} = left({text{subcond1}:1,0}{text{subcond2:1,0}}{text{subcond3:1,0}}=1right)$$
$$
y=left{\color{red}{left{x^2>1:1,0right}left{x^4<3:1,0right}=1:left(x^2-xright)},\color{blue}{left{x^2>1.8:1,0right}left{x^4+x^2<25:1,0right}=1:x},\color{green}{-1}right}$$
link
answered Nov 26 at 17:20
Vasily Mitch
1,32837
add a comment |
I haven't found a built-in conjunction function, so you might find helpful my workaround:
For simple domain restrictions, you can use a product of conditions:
$$y=f(x){text{cond1}}{text{cond2}}{text{cond3}}$$
$$y=(x^2-x)left{x^2>1right}left{x^4<3right}$$
link
For complex piecewise functions, you can use this hack:
$$text{cond1} = left({text{subcond1}:1,0}{text{subcond2:1,0}}{text{subcond3:1,0}}=1right)$$
$$
y=left{\color{red}{left{x^2>1:1,0right}left{x^4<3:1,0right}=1:left(x^2-xright)},\color{blue}{left{x^2>1.8:1,0right}left{x^4+x^2<25:1,0right}=1:x},\color{green}{-1}right}$$
link
answered Nov 26 at 17:20
Vasily Mitch
1,32837
add a comment |
I haven't found a built-in conjunction function, so you might find helpful my workaround:
For simple domain restrictions, you can use a product of conditions:
$$y=f(x){text{cond1}}{text{cond2}}{text{cond3}}$$
$$y=(x^2-x)left{x^2>1right}left{x^4<3right}$$
link
For complex piecewise functions, you can use this hack:
$$text{cond1} = left({text{subcond1}:1,0}{text{subcond2:1,0}}{text{subcond3:1,0}}=1right)$$
$$
y=left{\color{red}{left{x^2>1:1,0right}left{x^4<3:1,0right}=1:left(x^2-xright)},\color{blue}{left{x^2>1.8:1,0right}left{x^4+x^2<25:1,0right}=1:x},\color{green}{-1}right}$$
link
answered Nov 26 at 17:20
Vasily Mitch
1,32837
I haven't found a built-in conjunction function, so you might find helpful my workaround:
For simple domain restrictions, you can use a product of conditions:
$$y=f(x){text{cond1}}{text{cond2}}{text{cond3}}$$
$$y=(x^2-x)left{x^2>1right}left{x^4<3right}$$
link
For complex piecewise functions, you can use this hack:
$$text{cond1} = left({text{subcond1}:1,0}{text{subcond2:1,0}}{text{subcond3:1,0}}=1right)$$
$$
y=left{\color{red}{left{x^2>1:1,0right}left{x^4<3:1,0right}=1:left(x^2-xright)},\color{blue}{left{x^2>1.8:1,0right}left{x^4+x^2<25:1,0right}=1:x},\color{green}{-1}right}$$
link
answered Nov 26 at 17:20
Vasily Mitch
1,32837
answered Nov 26 at 17:20
Vasily Mitch
1,32837
answered Nov 26 at 17:20
Vasily Mitch
1,32837
answered Nov 26 at 17:20
Vasily Mitch
1,32837
1,32837
add a comment |
add a comment |
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