Open main menu
Home
Random
Recent changes
Special pages
Community portal
Preferences
About Wikipedia
Disclaimers
Incubator escapee wiki
Search
User menu
Talk
Dark mode
Contributions
Create account
Log in
Editing
Gradient
(section)
Warning:
You are not logged in. Your IP address will be publicly visible if you make any edits. If you
log in
or
create an account
, your edits will be attributed to your username, along with other benefits.
Anti-spam check. Do
not
fill this in!
===Gradient is direction of steepest ascent=== The gradient of a function <math>f \colon \R^n \to \R</math> at point {{math|''x''}} is also the direction of its steepest ascent, i.e. it maximizes its [[directional derivative]]: Let <math> v \in \R^n</math> be an arbitrary unit vector. With the directional derivative defined as <math display="block">\nabla_v f (x) = \lim_{h \rightarrow 0} \frac{f(x + vh) - f(x)}{h},</math> we get, by substituting the function <math>f(x + vh)</math> with its [[Taylor series]], <math display="block">\nabla_v f (x) = \lim_{h \rightarrow 0} \frac{(f(x) + \nabla f \cdot vh + R) - f(x)}{h},</math> where <math>R</math> denotes higher order terms in <math>vh</math>. Dividing by <math>h</math>, and taking the limit yields a term which is bounded from above by the [[Cauchy-Schwarz inequality]]<ref>{{cite book |author1=T. Arens | title=Mathematik |edition=5th |publisher=Springer Spektrum Berlin |year=2022 | doi=10.1007/978-3-662-64389-1 |isbn=978-3-662-64388-4 |url = https://doi.org/10.1007/978-3-662-64389-1}}</ref> <math display="block">|\nabla_v f (x)| = |\nabla f \cdot v| \le |\nabla f| |v| = |\nabla f|.</math> Choosing <math>v^* = \nabla f/|\nabla f|</math> maximizes the directional derivative, and equals the upper bound <math display="block">|\nabla_{v^*} f (x)| = |(\nabla f)^2/|\nabla f|| = |\nabla f|.</math>
Edit summary
(Briefly describe your changes)
By publishing changes, you agree to the
Terms of Use
, and you irrevocably agree to release your contribution under the
CC BY-SA 4.0 License
and the
GFDL
. You agree that a hyperlink or URL is sufficient attribution under the Creative Commons license.
Cancel
Editing help
(opens in new window)