Hi,
I first want to say, I'm kind of out of my depth here.
My apologies in advance if this is the wrong section, or I say something that is complete nonsense.
After much research experimenting with exponential decay, I found a website that used different models to fit curves to a specific dataset you give it. I found a specific formula that seems to fit my needs.
My problem is trying to understand the coefficients, V0 & K. How they're calculated.
Dataset:
K = 0.0063561076607244
V0 = 0.0159179044024703
So what I've found is the formula y(t) = a × ekt, which you can use to workout K. Which involves dividing by a, and taking the natrual log from both sides etc... but i just can't seem to get to K.
I also can't seem to find what V0 is. How it is calculated, I'm kind of going around in circles. I want to replicate this in Excel.
Is there anyone who can point me to some resources I need to learn to understand this, or maybe explain in a more intuitive way as to what these two coefficients are?
Thanks!
I first want to say, I'm kind of out of my depth here.
My apologies in advance if this is the wrong section, or I say something that is complete nonsense.
After much research experimenting with exponential decay, I found a website that used different models to fit curves to a specific dataset you give it. I found a specific formula that seems to fit my needs.
My problem is trying to understand the coefficients, V0 & K. How they're calculated.
Dataset:
X | Y |
0 | 6 |
500 | 3.6 |
1000 | 3.5 |
K = 0.0063561076607244
V0 = 0.0159179044024703
So what I've found is the formula y(t) = a × ekt, which you can use to workout K. Which involves dividing by a, and taking the natrual log from both sides etc... but i just can't seem to get to K.
I also can't seem to find what V0 is. How it is calculated, I'm kind of going around in circles. I want to replicate this in Excel.
Is there anyone who can point me to some resources I need to learn to understand this, or maybe explain in a more intuitive way as to what these two coefficients are?
Thanks!