"exterior derivative" meaning in All languages combined

{
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    {
      "form": "exterior derivatives",
      "tags": [
        "plural"
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      "expansion": "exterior derivative (plural exterior derivatives)",
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  "lang": "English",
  "lang_code": "en",
  "pos": "noun",
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      "examples": [
        {
          "text": "The exterior derivative of a “scalar”, i.e., a function f=f(x¹,x²,...,xⁿ) where the xⁱ’s are coordinates of ℝⁿ, is df=∂f/∂x¹dx¹+∂f/∂x²dx²+...+∂f/∂xⁿdxⁿ."
        },
        {
          "text": "The exterior derivative of a k-blade f,dx∧dx∧...∧dx is df∧dx∧dx∧...∧dx."
        },
        {
          "text": "The exterior derivative d may be though of as a differential operator del wedge: ∇∧, where ∇=∂/∂x¹dx¹+∂/∂x_2dx²+...+∂/∂xⁿdxⁿ. Then the square of the exterior derivative is d²=∇∧∇∧=(∇∧∇)∧=0∧=0 because the wedge product is alternating. (If u is a blade and f a scalar (function), then fu≡f∧u, so d(fu)=∇∧(fu)=∇∧(f∧u)=(∇∧f)∧u=df∧u.) Another way to show that d²=0 is that partial derivatives commute and wedge products of 1-forms anti-commute (so when d² is applied to a blade then the distributed parts end up canceling to zero.)"
        }
      ],
      "glosses": [
        "A differential operator which acts on a differential k-form to yield a differential (k+1)-form, unless the k-form is a pseudoscalar, in which case it yields 0."
      ],
      "id": "en-exterior_derivative-en-noun-D-oZNg1F",
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          "calculus",
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        [
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        ]
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      "raw_glosses": [
        "(calculus) A differential operator which acts on a differential k-form to yield a differential (k+1)-form, unless the k-form is a pseudoscalar, in which case it yields 0."
      ],
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        "calculus",
        "mathematics",
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      ],
      "wikipedia": [
        "exterior derivative"
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    }
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  "word": "exterior derivative"
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{
  "forms": [
    {
      "form": "exterior derivatives",
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      "examples": [
        {
          "text": "The exterior derivative of a “scalar”, i.e., a function f=f(x¹,x²,...,xⁿ) where the xⁱ’s are coordinates of ℝⁿ, is df=∂f/∂x¹dx¹+∂f/∂x²dx²+...+∂f/∂xⁿdxⁿ."
        },
        {
          "text": "The exterior derivative of a k-blade f,dx∧dx∧...∧dx is df∧dx∧dx∧...∧dx."
        },
        {
          "text": "The exterior derivative d may be though of as a differential operator del wedge: ∇∧, where ∇=∂/∂x¹dx¹+∂/∂x_2dx²+...+∂/∂xⁿdxⁿ. Then the square of the exterior derivative is d²=∇∧∇∧=(∇∧∇)∧=0∧=0 because the wedge product is alternating. (If u is a blade and f a scalar (function), then fu≡f∧u, so d(fu)=∇∧(fu)=∇∧(f∧u)=(∇∧f)∧u=df∧u.) Another way to show that d²=0 is that partial derivatives commute and wedge products of 1-forms anti-commute (so when d² is applied to a blade then the distributed parts end up canceling to zero.)"
        }
      ],
      "glosses": [
        "A differential operator which acts on a differential k-form to yield a differential (k+1)-form, unless the k-form is a pseudoscalar, in which case it yields 0."
      ],
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          "calculus",
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      "raw_glosses": [
        "(calculus) A differential operator which acts on a differential k-form to yield a differential (k+1)-form, unless the k-form is a pseudoscalar, in which case it yields 0."
      ],
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        "calculus",
        "mathematics",
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