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Korrektur von Peter Scheiblechner: Integral #84

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daniw opened this issue Jun 20, 2013 · 3 comments
Open

Korrektur von Peter Scheiblechner: Integral #84

daniw opened this issue Jun 20, 2013 · 3 comments
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@daniw
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daniw commented Jun 20, 2013

i=a+i\cdot\Delta x
@ninux
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ninux commented Jun 21, 2013

???

@daniw
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daniw commented Jun 21, 2013

Rückfrage an den Dozenten

@daniw
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daniw commented Jun 22, 2013

Antwort von Peter Scheiblechner:

ich habe mich vertan, ich meinte

x_i= a+i\cdot\Delta x.

Man bekommt die i-te Stützstelle, indem man i mal die Intervallbreite zu a addiert: x_0=a, x_1=a+\Delta x, x_2=a+2\Delta x etc.
Sie haben a vergessen und nur \Delat statt \Delta x geschrieben.

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