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Sin

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[SOUND].

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It's not too painful to differentiate sin,
even if

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we gotta do it a whole bunch of times.

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Well, here we go.

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here's my function.
I'm going to call it f.

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So f of x is sin x.
And now let's start differentiating.

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So, the derivative of f, derivative of
sine, is cosine.

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The derivative of cosine is negative sine,

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so that 's the second derivative sine.

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The derivative of negative sine is
negative

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cosine, so it's the third derivative sine.

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The fourth derivative of F is the

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derivative of negative cosine which is,
sine again.

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because it's negative, negative sine so
it's just sine.

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The fifth derivative of sine is the
derivative of sine again, which is just

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cosine of X again.

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Now, let's evaluate those derivatives at
zero.

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Well, f of zero is zero.

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F prime of zero cosine of zero is one.
F double prime of zero.

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So what's negative sine of zero?
That's zero.

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The third derivative at zero is negative
cosine of zero.

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That's negative one.

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The fourth derivative at zero is sine of
zero.

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That's zero.

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The fifth derivative at zero is cosine of
zero, which is one.

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We're seeing a pattern.

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You know what's really going on, is that
the fourth derivative of sine is itself.

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So the fifth derivative of sine, is the
same as the first derivative of sine.

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Means the sixth derivative of sine is the
same as the second derivative of sine.

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So, if we're just looking at these
derivatives at zero,

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we're seeing 0, 1, 0, minus 1 0, 1.
What's the sixth derivative at zero?

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It's going to be zero again.

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What's the seventh derivative at zero?
Well.

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It's got to be minus 1.

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Now because it's 0,1,0 minus 1 and fourth
derivative

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is the same as the 0 of the derivative.

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Like the fourth derivative is the function
itself again.

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So, it's 0,1,0 minus 1,

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0,1,0 minus 1 that 8th derivative at 0

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would be 0, the 9th derivative at 0 would
be 1.

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The tenth derivative at zero would be
zero.

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The eleventh derivative at zero would be
minus 1.

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And they just keep on going.

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0, 1, 0, minus 1, 0, and so on, and so
forth.

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Finding that pattern

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lets us write down the Taylor series
around zero.

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Well, here's what it ends up being.

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Or at least this is one way of writing
down what it ends up being.

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So, the Taylor series around zero.

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So, centered around zero.
For Sine is the following.

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It's the sum, angles from zero to infinity
of negative one to the nth power

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divided by 2n plus 1 factorial times X to
the 2n plus one power but why does

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that work?

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When you think about this, I've made this
table here.

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So, this is our little table that shows n
and in

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below here is the nth derivative of Sine
evaluated at 0.

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I'd say the 0th derivative is just a
function Sine at 0.

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The first derivative is Cosine at 0 which
is 1.

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So, am using the numbers that we have
computed before and we've

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got this pattern that we saw 0, 1, 0,
minus 1, 0,

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1, 0, minus 1 and so on.

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But we should notice Is that in all the
even derivatives are 0.

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So this thing doesn't include any terms
with x to an

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even power, and that's why I've written x
to 2n plus 1.

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This is kind of a sneaky trick just for
recording all of the odd powers.

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And what about those terms with x to an
odd power?

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Well, when n is an odd number, I'm
flip-flopping in sign,

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in sign, and that's exactly what this
minus 1 to the n accomplishes for me.

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This is all pretty great.

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But it's raising a very serious question.

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Is this power series actually equal to sin
of x on any interval?

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That is the problem.

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We're assuming that sine has a power

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series representation valid on an interval
around zero.

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And if it can be represented as a power
series, then we've

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figured out what the power series has to
be.

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But how do we know that sine has any power
series representation?

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That's the problem that remains.

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[NOISE]