### Video Transcript

If the Maclaurin series of the
function π is π of π₯ is equal to two minus one over six multiplied by π₯ plus
five divided by 24 multiplied by π₯ squared minus seven divided by 60 multiplied by
π₯ cubed plus three divided by 80 multiplied by π₯ to the fourth power and so
on. Find the equation of the tangent to
the curve of π at π₯ is equal to zero.

The question gives us the Maclaurin
series for a function π and asks us to find the equation of the tangent to the
curve of π when π₯ is equal to zero. We recall that a straight line will
have equation π¦ is equal to ππ₯ plus π where π is the slope of our line and π
is the π¦-intercept. We also recall that our function π
prime of π₯ gives us the slope of the tangent to the curve of π at π₯. And since the question asks us to
find the tangent when π₯ is equal to zero, we can conclude that the slope of our
tangent is given by π prime of zero.

Now, we recall that the Maclaurin
series of a function is equal to the Taylor series expansion about zero. Namely, this gives us that π of π₯
is equal to π evaluated at zero plus the first derivative of π evaluated at zero
multiplied by π₯ plus the second derivative of π evaluated at zero divided by two
factorial multiplied by π₯ squared plus the third derivative of π evaluated at zero
divided by three factorial multiplied by π₯ cubed. And if we keep going and going,
then we get the Maclaurin series for our function π.

From the question weβre given the
Maclaurin series of our function π is equal to two plus negative one-sixth
multiplied by π₯ plus five over 24 multiplied by π₯ squared plus negative seven
divided by 60 multiplied by π₯ cubed. And this keeps going and going. We can see from the definition of
our Maclaurin series that the coefficient of π₯ is equal to π prime of zero. Therefore, we must have that the
first derivative of π evaluated at zero is equal to negative one-sixth. So we have the slope of our tangent
is negative one-sixth giving us the equation π¦ is equal to negative one-sixth
multiplied by π₯ plus some constant π.

One way to find the value of π is
to find the π₯- and π¦-coordinates of a point on our tangent line. And one way to do this is to notice
that since weβre finding the tangent line to the curve of π when π₯ is equal to
zero, we must have that the tangent line and the curve of π meet when π₯ is equal
to zero. Therefore, we must have that when
π₯ is equal to zero, our π¦-value is equal to the function π evaluated at zero. Now, we just need to find the value
of the function π evaluated at zero. And we can do this by looking on
our Maclaurin series. We can see that the first term in
our series is equal to the function π evaluated at zero. So we must have that π zero is
equal to two.

So what we have just shown is that
the 0.2 is on our tangent line. Therefore, if we substitute the
value of π₯ equals zero and the value of π¦ equals two into the equation for our
line, we get two is equal to negative one-sixth multiplied by zero plus π. And since negative one-sixth
multiplied by zero is just equal to zero, we must have that π is equal to two. So what weβve shown is that for the
Maclaurin series π of π₯ given to us in the question, the equation of the tangent
to the curve π at π₯ is equal to zero is π¦ is equal to negative one-sixth π₯ plus
two.