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JingleBells

  • 3 years ago

In Video lecture 05 round about 37' where Pr. Jerison draws the f(x)=tan(x) and f(x)=arctan(x) the two curve crosses where if I plug in real data I can't get the same results, did he lowered the curve to just to illustrate the symmetry along the y=x line or do you have to do some kind of stretch to get the same picture?

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  1. keshavsarda
    • 3 years ago
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    both are inverse function.........so they are symmetric to line y=x

  2. JingleBells
    • 3 years ago
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    Thanks, that bit I get, what I'm not sure about is that does the graph of tanx and arctanx overlap or not.

  3. keshavsarda
    • 3 years ago
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    yes they meet at y=x=1

  4. JingleBells
    • 3 years ago
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    So do you know what I could do to get the same graph as in the video where the two curves cross each other?

  5. keshavsarda
    • 3 years ago
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    no, u cant draw that graps....because we are not accurate. But u can roughly sketch it

  6. JingleBells
    • 3 years ago
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    I have quick graph on my iPhone and the two graphs never cross that's why I'm asking :-)

  7. keshavsarda
    • 3 years ago
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    |dw:1340852698875:dw|

  8. JingleBells
    • 3 years ago
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    Thanks but if you plotted the two graphs with real values of tan(x) and arctan(x) they don't cross at all.

  9. kutulu
    • 3 years ago
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    if you watch the recitation video for Session 15, it addresses your question. The profession drew the graph of arctan wrong. Within the single domain of -pi/2 -> pi/2, the only place that tan and arctan cross is at x=0. Arctan should be *below* y=x everywhere else while tan should be *above* y=x everywhere else.

  10. kutulu
    • 3 years ago
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    erm. make that The professor* drew the graph...

  11. JingleBells
    • 3 years ago
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    Thanks a million! I'm using iTunes U to watch the videos and I will, from now, watch the recitations on the web! :-)

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