How to apply a function to successive overlapping list elements?

Question

I have a list of elements and I would like to create a new list by applying a function to successive overlapping pairs from the original list.

Specifically, I am trying to compute the Pythagorean sum of each pair. So, if the original list is {$a,b,c,d,...$} I want {$\sqrt{a^2+b^2},\sqrt{b^2+c^2},\sqrt{c^2+d^2},...$}

Since this is similar to the built-in Differences function, (which turns the original list into {$b-a,c-b,d-c,...$}), I expected to find a built-in function along the lines of BuiltIn[f,{a,b,c,d,...},options] where f is defined by the user. But alas, my search has been fruitless...

Thanks in advance!

-----------EDIT-----------

Right after posting this, I came up with

Sqrt[#1^2 + #2^2] & @@@ Subsequences[#, {2}] &@{a, b, c, d}

which does the trick. I was going to update that here, but you all already came through with your answers, so I'll try those as well. Thanks!

Answer 1

You can use BlockMap + Norm

BlockMap[Norm, {a, b, c, d, e}, 2, 1]

{Sqrt[Abs[a]^2 + Abs[b]^2], Sqrt[Abs[b]^2 + Abs[c]^2], 
  Sqrt[Abs[c]^2 + Abs[d]^2], Sqrt[Abs[d]^2 + Abs[e]^2]}

TeXForm @ Simplify[%, Thread[{a, b, c, d, e} >= 0]]

$\left\{\sqrt{a^2+b^2},\sqrt{b^2+c^2},\sqrt{c^2+d^2},\sqrt{d^2+e^2}\right\}$

Answer 2

list = {a, b, c, d, e};

Using SequenceCases

SequenceCases[list, x :{_, _} :> Norm[x], Overlaps -> True]

Answer 3

l = {a, b, c, d, e};

---

Using FoldPairList:

f = List /@ FoldPairList[TakeDrop, #, {2, 2}] & /@ {#, RotateLeft@#} &;

Simplify[Norm @@@ Riffle @@ f@l, Thread[l > 0]]

A simpler alternative that uses FoldPairList (Thanks to @Syed!):

FoldPairList[ {Sqrt[#1^2 + #2^2], #2} &, list]