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The idea of the gradient of a function is to ask this question: When I jiggle the input, how much does the output change? That can depend on where you're doin

The idea of the gradient of a function is to ask this question: When I jiggle the input, how much does the output change? That can depend on where you're doing it. For f(x)=x^2, jiggling x around 0 results in almost no change at all: f(-0.1) = 0.01 f( 0.0) = 0.00 f( 0.1) = 0.01 But jiggling around a value of 10 makes a much larger difference: f( 9.9) = 98.01 f(10.0) = 100.00 f(10.1) = 102.01 Here, around the value of x=10, jiggling x up and down by 0.1 changes the output by about 2. Experimenting a lot, you can discover that for f(x)=x^2 you will find that when the jiggling (call it j) is small, around a value x, the amount of change is roughly (2x)(j). So roughly: f(x+j) ~ f(x)+2xj Gradient is usually expressed as "Rise over Run". The "Rise" here is the change in f, so it's f(x+j)-f(x), and the "Run" is j, the amount of jiggle, and what this has shown is that for f(x)=x^2, "Rise of Run", the gradient, is (roughly) 2x. You can write all that in code, but those are the ideas, and I hope it answers your question.

coolgeek 5y ago
> you can discover that for f(x)=x^2 you will find that when the jiggling (call it j) is small, around a value x, the amount of change is roughly (2x)(j) Are you implying that the change corresponds to the derivative of the function, or is that just a coincidence in this case? (I'm guessing from your characterization of rise over run that the answer is yes. But you know what they say about assumptions.)
- ColinWright 5y ago
  >> you can discover that for f(x)=x^2 you will find that when the jiggling (call it j) is small, around a value x, the amount of change is roughly (2x)(j) > Are you implying that the change corresponds to the derivative of the function, or is that just a coincidence in this case? When you have a function f(x), the amount that f(x) changes as you jiggle the value x is roughly given by the amount of jiggle times the derivative f'(x) [also known by many other names]. That's more-or-less what "derivative" means in this context. So under appropriate and rather generous conditions: f(x+j) ~ f(x) + j.f'(x) Does that answer your question? I've answered the question I think you asked, but that's not always the the question you intended.
  - coolgeek 5y ago
    I guess the question I was really asking was is the gradient of a function equivalent to the derivative of the function? But a quick look at the wiki page for gradient suggests that my calc I/II intuition of derivatives is insufficient here
    - ColinWright 5y ago
      For reasonable functions f:R->R, yes, the gradient of the graph of the function is the value of the derivative at that point. But "Gradient" is a term that's used in wider contexts, so the real answer is "it depends". But I've had a look at this page: https://en.wikipedia.org/wiki/Gradient https://en.wikipedia.org/wiki/Gradient And that really does say that the "Gradient" is basically the derivative, suitably generalised for larger dimensions.

coolgeek 5y ago

> you can discover that for f(x)=x^2 you will find that when the jiggling (call it j) is small, around a value x, the amount of change is roughly (2x)(j) Are you implying that the change corresponds to the derivative of the function, or is that just a coincidence in this case? (I'm guessing from your characterization of rise over run that the answer is yes. But you know what they say about assumptions.)

ColinWright 5y ago
>> you can discover that for f(x)=x^2 you will find that when the jiggling (call it j) is small, around a value x, the amount of change is roughly (2x)(j) > Are you implying that the change corresponds to the derivative of the function, or is that just a coincidence in this case? When you have a function f(x), the amount that f(x) changes as you jiggle the value x is roughly given by the amount of jiggle times the derivative f'(x) [also known by many other names]. That's more-or-less what "derivative" means in this context. So under appropriate and rather generous conditions: f(x+j) ~ f(x) + j.f'(x) Does that answer your question? I've answered the question I think you asked, but that's not always the the question you intended.
- coolgeek 5y ago
  I guess the question I was really asking was is the gradient of a function equivalent to the derivative of the function? But a quick look at the wiki page for gradient suggests that my calc I/II intuition of derivatives is insufficient here
  - ColinWright 5y ago
    For reasonable functions f:R->R, yes, the gradient of the graph of the function is the value of the derivative at that point. But "Gradient" is a term that's used in wider contexts, so the real answer is "it depends". But I've had a look at this page: https://en.wikipedia.org/wiki/Gradient https://en.wikipedia.org/wiki/Gradient And that really does say that the "Gradient" is basically the derivative, suitably generalised for larger dimensions.