How to get Y value from exponential function that starts at 0,0 and ends at 100,1000
up vote
2
down vote
favorite
It's been a while since my math classes so here's the question, any help is appreciated.
I'll be implementing the function in javascript also.
I have an X range of 0-100 and a set max Y of 10000 (which may change).
This is an exponential function and the closer X is to 100, the closer it gets to Y max and x=100 is y=10000. So my question is, what is the function that can handle this? Ideally I'd like to have something to control the steepness of the curve as well.
What I have in mind :
graph
Thanks.
javascript function math graph linear-algebra
asked Nov 10 at 22:11
LAZ
385
|
show 1 more comment
up vote
2
down vote
favorite
It's been a while since my math classes so here's the question, any help is appreciated.
I'll be implementing the function in javascript also.
I have an X range of 0-100 and a set max Y of 10000 (which may change).
This is an exponential function and the closer X is to 100, the closer it gets to Y max and x=100 is y=10000. So my question is, what is the function that can handle this? Ideally I'd like to have something to control the steepness of the curve as well.
What I have in mind :
graph
Thanks.
javascript function math graph linear-algebra
asked Nov 10 at 22:11
LAZ
385
Do you want to know how to get the Y value in javascript?
– Donny Verduijn
Nov 10 at 22:36
Yeah, or just the expression, I can manage putting it in js from there. Whatever is easiest for you.
– LAZ
Nov 10 at 22:38
1
Because otherwise, this question should be asked on math.stackexchange.com
– Donny Verduijn
Nov 10 at 22:40
Well it will be used in javascript
– LAZ
Nov 10 at 22:43
Programming goes hand to wand with maths, thus this is really just a math question... this indeed should be asked on math.stackexchange.com
– Akxe
Nov 11 at 1:57
|
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
It's been a while since my math classes so here's the question, any help is appreciated.
I'll be implementing the function in javascript also.
I have an X range of 0-100 and a set max Y of 10000 (which may change).
This is an exponential function and the closer X is to 100, the closer it gets to Y max and x=100 is y=10000. So my question is, what is the function that can handle this? Ideally I'd like to have something to control the steepness of the curve as well.
What I have in mind :
graph
Thanks.
javascript function math graph linear-algebra
asked Nov 10 at 22:11
LAZ
385
It's been a while since my math classes so here's the question, any help is appreciated.
I'll be implementing the function in javascript also.
I have an X range of 0-100 and a set max Y of 10000 (which may change).
This is an exponential function and the closer X is to 100, the closer it gets to Y max and x=100 is y=10000. So my question is, what is the function that can handle this? Ideally I'd like to have something to control the steepness of the curve as well.
What I have in mind :
graph
Thanks.
javascript function math graph linear-algebra
javascript function math graph linear-algebra
asked Nov 10 at 22:11
LAZ
385
asked Nov 10 at 22:11
LAZ
385
edited Nov 10 at 22:27
asked Nov 10 at 22:11
LAZ
385
asked Nov 10 at 22:11
LAZ
385
asked Nov 10 at 22:11
LAZ
385
385
Do you want to know how to get the Y value in javascript?
– Donny Verduijn
Nov 10 at 22:36
Yeah, or just the expression, I can manage putting it in js from there. Whatever is easiest for you.
– LAZ
Nov 10 at 22:38
1
Because otherwise, this question should be asked on math.stackexchange.com
– Donny Verduijn
Nov 10 at 22:40
Well it will be used in javascript
– LAZ
Nov 10 at 22:43
Programming goes hand to wand with maths, thus this is really just a math question... this indeed should be asked on math.stackexchange.com
– Akxe
Nov 11 at 1:57
|
show 1 more comment
Do you want to know how to get the Y value in javascript?
– Donny Verduijn
Nov 10 at 22:36
Yeah, or just the expression, I can manage putting it in js from there. Whatever is easiest for you.
– LAZ
Nov 10 at 22:38
1
Because otherwise, this question should be asked on math.stackexchange.com
– Donny Verduijn
Nov 10 at 22:40
Well it will be used in javascript
– LAZ
Nov 10 at 22:43
Programming goes hand to wand with maths, thus this is really just a math question... this indeed should be asked on math.stackexchange.com
– Akxe
Nov 11 at 1:57
Do you want to know how to get the Y value in javascript?
– Donny Verduijn
Nov 10 at 22:36
Do you want to know how to get the Y value in javascript?
– Donny Verduijn
Nov 10 at 22:36
Yeah, or just the expression, I can manage putting it in js from there. Whatever is easiest for you.
– LAZ
Nov 10 at 22:38
Yeah, or just the expression, I can manage putting it in js from there. Whatever is easiest for you.
– LAZ
Nov 10 at 22:38
1
1
Because otherwise, this question should be asked on math.stackexchange.com
– Donny Verduijn
Nov 10 at 22:40
Because otherwise, this question should be asked on math.stackexchange.com
– Donny Verduijn
Nov 10 at 22:40
Well it will be used in javascript
– LAZ
Nov 10 at 22:43
Well it will be used in javascript
– LAZ
Nov 10 at 22:43
Programming goes hand to wand with maths, thus this is really just a math question... this indeed should be asked on math.stackexchange.com
– Akxe
Nov 11 at 1:57
Programming goes hand to wand with maths, thus this is really just a math question... this indeed should be asked on math.stackexchange.com
– Akxe
Nov 11 at 1:57
|
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
The formula would be a variant of y = a * exp(b * x), where you can play around with a and b to get the curve / endpoints you want. For 100/100.000, you could use:
y = 5 * exp(0.0990348754944493 * x). In Javascript that would look something like:
function getExponent(x) {
return 5 * Math.pow(0.0990348754944493*x);
}
or if you want a and b to be variables:
function getExponent(x,a,b) {
return a * Math.pow(b*x);
}
answered Nov 11 at 1:48
CIAndrews
1766
I want to be able to provide x and the y value at x=100. I can't go in the code and edit the floats every time I have a different y max. For example, what is Y when X = 56 when the exponential starts at (0,0) and intercepts with the point (100,10000)? The range of 0-100 will never change, the steepness I will play with but eventually will be static and the "max" Y value will vary. Think of it like a slider, you slide it up to the max which in this case is 10000. But some sliders go up to 100000 or a million, etc.
– LAZ
Nov 11 at 2:53
If you want to make the steepness static, we can assume b is (becomes) fixed. Since y_max and x_max are known (although y_max is varying), you can use a function to calculate a: a = exp(b * x_max) / y_max = exp(b * 100) / y_max Then... whenever you have fixed b, the interceptions (0,0) and (100, y_max) will always hold.
– CIAndrews
Nov 11 at 3:20
Thanks, makes sense but Math.pow() requires two args, and exp() doesn't exist in my environment. I did try Math.exp(b * x_max) / y_max but that returns an 18 digit number.
– LAZ
Nov 11 at 3:53
Ah yes, my b was incorrect. Changed it to 0.0990348754944493, that results exactly in y = 100.000 for x = 100 and a = 5
– CIAndrews
Nov 11 at 4:18
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The formula would be a variant of y = a * exp(b * x), where you can play around with a and b to get the curve / endpoints you want. For 100/100.000, you could use:
y = 5 * exp(0.0990348754944493 * x). In Javascript that would look something like:
function getExponent(x) {
return 5 * Math.pow(0.0990348754944493*x);
}
or if you want a and b to be variables:
function getExponent(x,a,b) {
return a * Math.pow(b*x);
}
answered Nov 11 at 1:48
CIAndrews
1766
I want to be able to provide x and the y value at x=100. I can't go in the code and edit the floats every time I have a different y max. For example, what is Y when X = 56 when the exponential starts at (0,0) and intercepts with the point (100,10000)? The range of 0-100 will never change, the steepness I will play with but eventually will be static and the "max" Y value will vary. Think of it like a slider, you slide it up to the max which in this case is 10000. But some sliders go up to 100000 or a million, etc.
– LAZ
Nov 11 at 2:53
If you want to make the steepness static, we can assume b is (becomes) fixed. Since y_max and x_max are known (although y_max is varying), you can use a function to calculate a: a = exp(b * x_max) / y_max = exp(b * 100) / y_max Then... whenever you have fixed b, the interceptions (0,0) and (100, y_max) will always hold.
– CIAndrews
Nov 11 at 3:20
Thanks, makes sense but Math.pow() requires two args, and exp() doesn't exist in my environment. I did try Math.exp(b * x_max) / y_max but that returns an 18 digit number.
– LAZ
Nov 11 at 3:53
Ah yes, my b was incorrect. Changed it to 0.0990348754944493, that results exactly in y = 100.000 for x = 100 and a = 5
– CIAndrews
Nov 11 at 4:18
add a comment |
up vote
1
down vote
accepted
The formula would be a variant of y = a * exp(b * x), where you can play around with a and b to get the curve / endpoints you want. For 100/100.000, you could use:
y = 5 * exp(0.0990348754944493 * x). In Javascript that would look something like:
function getExponent(x) {
return 5 * Math.pow(0.0990348754944493*x);
}
or if you want a and b to be variables:
function getExponent(x,a,b) {
return a * Math.pow(b*x);
}
answered Nov 11 at 1:48
CIAndrews
1766
I want to be able to provide x and the y value at x=100. I can't go in the code and edit the floats every time I have a different y max. For example, what is Y when X = 56 when the exponential starts at (0,0) and intercepts with the point (100,10000)? The range of 0-100 will never change, the steepness I will play with but eventually will be static and the "max" Y value will vary. Think of it like a slider, you slide it up to the max which in this case is 10000. But some sliders go up to 100000 or a million, etc.
– LAZ
Nov 11 at 2:53
If you want to make the steepness static, we can assume b is (becomes) fixed. Since y_max and x_max are known (although y_max is varying), you can use a function to calculate a: a = exp(b * x_max) / y_max = exp(b * 100) / y_max Then... whenever you have fixed b, the interceptions (0,0) and (100, y_max) will always hold.
– CIAndrews
Nov 11 at 3:20
Thanks, makes sense but Math.pow() requires two args, and exp() doesn't exist in my environment. I did try Math.exp(b * x_max) / y_max but that returns an 18 digit number.
– LAZ
Nov 11 at 3:53
Ah yes, my b was incorrect. Changed it to 0.0990348754944493, that results exactly in y = 100.000 for x = 100 and a = 5
– CIAndrews
Nov 11 at 4:18
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The formula would be a variant of y = a * exp(b * x), where you can play around with a and b to get the curve / endpoints you want. For 100/100.000, you could use:
y = 5 * exp(0.0990348754944493 * x). In Javascript that would look something like:
function getExponent(x) {
return 5 * Math.pow(0.0990348754944493*x);
}
or if you want a and b to be variables:
function getExponent(x,a,b) {
return a * Math.pow(b*x);
}
answered Nov 11 at 1:48
CIAndrews
1766
The formula would be a variant of y = a * exp(b * x), where you can play around with a and b to get the curve / endpoints you want. For 100/100.000, you could use:
y = 5 * exp(0.0990348754944493 * x). In Javascript that would look something like:
function getExponent(x) {
return 5 * Math.pow(0.0990348754944493*x);
}
or if you want a and b to be variables:
function getExponent(x,a,b) {
return a * Math.pow(b*x);
}
answered Nov 11 at 1:48
CIAndrews
1766
edited Nov 11 at 4:18
answered Nov 11 at 1:48
CIAndrews
1766
answered Nov 11 at 1:48
CIAndrews
1766
answered Nov 11 at 1:48
CIAndrews
1766
1766
I want to be able to provide x and the y value at x=100. I can't go in the code and edit the floats every time I have a different y max. For example, what is Y when X = 56 when the exponential starts at (0,0) and intercepts with the point (100,10000)? The range of 0-100 will never change, the steepness I will play with but eventually will be static and the "max" Y value will vary. Think of it like a slider, you slide it up to the max which in this case is 10000. But some sliders go up to 100000 or a million, etc.
– LAZ
Nov 11 at 2:53
If you want to make the steepness static, we can assume b is (becomes) fixed. Since y_max and x_max are known (although y_max is varying), you can use a function to calculate a: a = exp(b * x_max) / y_max = exp(b * 100) / y_max Then... whenever you have fixed b, the interceptions (0,0) and (100, y_max) will always hold.
– CIAndrews
Nov 11 at 3:20
Thanks, makes sense but Math.pow() requires two args, and exp() doesn't exist in my environment. I did try Math.exp(b * x_max) / y_max but that returns an 18 digit number.
– LAZ
Nov 11 at 3:53
Ah yes, my b was incorrect. Changed it to 0.0990348754944493, that results exactly in y = 100.000 for x = 100 and a = 5
– CIAndrews
Nov 11 at 4:18
add a comment |
I want to be able to provide x and the y value at x=100. I can't go in the code and edit the floats every time I have a different y max. For example, what is Y when X = 56 when the exponential starts at (0,0) and intercepts with the point (100,10000)? The range of 0-100 will never change, the steepness I will play with but eventually will be static and the "max" Y value will vary. Think of it like a slider, you slide it up to the max which in this case is 10000. But some sliders go up to 100000 or a million, etc.
– LAZ
Nov 11 at 2:53
If you want to make the steepness static, we can assume b is (becomes) fixed. Since y_max and x_max are known (although y_max is varying), you can use a function to calculate a: a = exp(b * x_max) / y_max = exp(b * 100) / y_max Then... whenever you have fixed b, the interceptions (0,0) and (100, y_max) will always hold.
– CIAndrews
Nov 11 at 3:20
Thanks, makes sense but Math.pow() requires two args, and exp() doesn't exist in my environment. I did try Math.exp(b * x_max) / y_max but that returns an 18 digit number.
– LAZ
Nov 11 at 3:53
Ah yes, my b was incorrect. Changed it to 0.0990348754944493, that results exactly in y = 100.000 for x = 100 and a = 5
– CIAndrews
Nov 11 at 4:18
I want to be able to provide x and the y value at x=100. I can't go in the code and edit the floats every time I have a different y max. For example, what is Y when X = 56 when the exponential starts at (0,0) and intercepts with the point (100,10000)? The range of 0-100 will never change, the steepness I will play with but eventually will be static and the "max" Y value will vary. Think of it like a slider, you slide it up to the max which in this case is 10000. But some sliders go up to 100000 or a million, etc.
– LAZ
Nov 11 at 2:53
I want to be able to provide x and the y value at x=100. I can't go in the code and edit the floats every time I have a different y max. For example, what is Y when X = 56 when the exponential starts at (0,0) and intercepts with the point (100,10000)? The range of 0-100 will never change, the steepness I will play with but eventually will be static and the "max" Y value will vary. Think of it like a slider, you slide it up to the max which in this case is 10000. But some sliders go up to 100000 or a million, etc.
– LAZ
Nov 11 at 2:53
If you want to make the steepness static, we can assume b is (becomes) fixed. Since y_max and x_max are known (although y_max is varying), you can use a function to calculate a: a = exp(b * x_max) / y_max = exp(b * 100) / y_max Then... whenever you have fixed b, the interceptions (0,0) and (100, y_max) will always hold.
– CIAndrews
Nov 11 at 3:20
If you want to make the steepness static, we can assume b is (becomes) fixed. Since y_max and x_max are known (although y_max is varying), you can use a function to calculate a: a = exp(b * x_max) / y_max = exp(b * 100) / y_max Then... whenever you have fixed b, the interceptions (0,0) and (100, y_max) will always hold.
– CIAndrews
Nov 11 at 3:20
Thanks, makes sense but Math.pow() requires two args, and exp() doesn't exist in my environment. I did try Math.exp(b * x_max) / y_max but that returns an 18 digit number.
– LAZ
Nov 11 at 3:53
Thanks, makes sense but Math.pow() requires two args, and exp() doesn't exist in my environment. I did try Math.exp(b * x_max) / y_max but that returns an 18 digit number.
– LAZ
Nov 11 at 3:53
Ah yes, my b was incorrect. Changed it to 0.0990348754944493, that results exactly in y = 100.000 for x = 100 and a = 5
– CIAndrews
Nov 11 at 4:18
Ah yes, my b was incorrect. Changed it to 0.0990348754944493, that results exactly in y = 100.000 for x = 100 and a = 5
– CIAndrews
Nov 11 at 4:18
add a comment |
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Do you want to know how to get the Y value in javascript?
– Donny Verduijn
Nov 10 at 22:36
Yeah, or just the expression, I can manage putting it in js from there. Whatever is easiest for you.
– LAZ
Nov 10 at 22:38
1
Because otherwise, this question should be asked on math.stackexchange.com
– Donny Verduijn
Nov 10 at 22:40
Well it will be used in javascript
– LAZ
Nov 10 at 22:43
Programming goes hand to wand with maths, thus this is really just a math question... this indeed should be asked on math.stackexchange.com
– Akxe
Nov 11 at 1:57