Math Is Fun Forum

Has anyone else learned about Maclaurin polynomials? I was suprised to read about it, becuase it allows you to fnd the formula's for sine, cosine, e^x and many other calculator functions. A while back I was asking around for the formula's for sine, cosine and tangent, and hardly anyone knew. Thats why I'm curious if many other calculus books study these.

The maclaurin polynomial of f(x)

f(0)/0! + f'(0)(x^1)/1! +  f''(0)( x^2 ) /2! + f'''(0)( x^3 )/3! ...

note you begin by multiplying the first term by f(0), you take the first derivative, evaluate it at zero, and multiply that by the second term, you take the third derivative, evaluate it at zero, and multiply it by the third term, and so on.

Supposedly, this creates a sort of replica of the function, but in the form of the polynomial. If the polynomial does not terminate, the accuracy of the function depends on how many terms you choose to use.

If you have a polynomial like 3x^2 + 2x + 6

f(x) = 3x^2 + 2x +6   so   f(0) = 6

f'(x) = 6x + 2             so   f'(0) = 2

f''(x) = 6                    so  f''(0) = 6

The third derivative and anything beyond that will be 0 so we can stop here.

Now lets evaluate the maclaurin polynomial for the function:

f(0)/0! + f'(0)(x^1)/1! +  f''(0)( x^2 ) /2! + f'''(0)( x^3 )/3! ...

= 6/0! + 2x/1! + (6x^2)/2!

= 6 + 2x + 3x^2

We found the maclaurin polynomail of 3x^2 +2x +6 and got 3x^2 + 2x + 6! From this we see that the maclaurin polynomail of a polynomial, is the polynomial itself!

Now take e^x

f (x) = e^x      f(0) = 1

f'(x) = e^x      f''(0) = 1

f''(x) = e^x    f''(0) = 1

Obviously the first second third and nth derivative of e^x will always be e^x, evaluated at zero, they are one, so if write the maclaurin polynomail:

f(0)/0! + f'(0)(x^1)/1! +  f''(0)( x^2 ) /2! + f'''(0)( x^3 )/3! ...

We have:

x^0/0! + x^1 + (x^2)/2! + (x^3)/3! + (x^4)/4! ....

and so on! Note the first term could have just been written as 1, I just wrote it that way to show the pattern is consistant.

So now we have a polynomial formula for e^x! :-D

You can also use this to find the formula's for sine and cosine, and other trigonometric functions.

I'm not sure how to find the formula for a function such as y = ln x because ln 0 does not exist. But maybe I'll learn more later.

The problems in my book often give me a functon and ask me to find the maclaurin polynomial and write the answer in summation notation. Writing them in summation notation can be a tricky buisiness at times, becuase the signes alternate, some terms disappear when their coefficiants are zero, and sometimes the factorials cancel eachother out. Its tricky but they're kind of fun.

I had understood that  7 ln 7 = ln 6^7   And that ln 6 - ln 2 = ln 6/2  (or 3)

But if you have:

7 ln 6 - ln 3

and you first write it as 7 ln 6/3 you have:

7 ln 2

If you first rewrite it as

ln 6^7 - ln 3

then you have:

ln (6^7)/3

But this produces a different answer then 7 ln 2

This seems to indicate you should use the power/coefficent rule last, but what if you had:

5 ln 3 - 4 ln 2

you'd have to use the power coefficiant rule first.

I think I'm just forgetting a few things. I haven't done logarithmic equations in while.

Find the angle between the slopes of the graphs y = x^3 and y = sqrt(x)

First we find the intersection(s) of the two functions.

x^3 = x^(1/2)

divide both sides by x^(1/2)

x^5/2 = 1

x = 1         took the 5/2th root of both sides.

If we find the derivatives of the original functions, we get:

3x^2 and -1/2sqrt(x)

Evaluated at the intersection (1), the slopes of the two lines are:

3 and -0.5

The angle between the slopes of two lines is given by:

tan A = (M1 - M2)/(1 - M1M2)

If we use 3 for M1 and -0.5 for M2 it evaluates to (5/2)/(5/2) which is 1.

So if the tangent of A equals 1, we know it must be either 45 or 225. 225 makes no sense so it must be 45. But in my book, the answer is 45 and 90. Ninety?! Where did that come from? If anything, I would have thought 45 or 135. (since two intersecting lines will form two pairs of suplementary angles) but 90 makes no sense at all.

What bugs me about the forumula  tan A = (M1 - M2)/(1 - M1M2) is that it makes no specifications about which is M1 and which is M2. I had two slopes of 3 and -0.5. Depending on which I choose for M1 and M2 changes the sign of the outcome of the formula. I'll either get 1 or -1.

Dismaying.

I was solving an equation for velocity, by differentiating the equation of position. I had to find the values of the f(x) for which the velocity is zero, greater then zero, or less then zero. But the velocity equation was a quadratic. So the velocity was zero at two points. But the times when the velocity was greater or less then zero, thats not so simple. The graph of the quadratic will be a parabolla that oppens either upward or downard. It will cross the x axis at the zero's of the equation. Now if it opens upward, then the points between the two zero's will be below the x axis. If it opens downward, the points between the two zero's will be above the x axis.

We can find the region between the zero's by simply designating it to be the region less then the larg zero and greater then the small zero. And the region outside the zero's by the points that are either greater then the large or less then the small.

So all we need to do is find out whether the parabolla opens upward or downward. Thats easy, if the equation is ax^2 + bx + c, then if a is positive, it will be large and positve for large absolute values of x (opens upward). If a is negative, then it will be large and negative for large absolute values of x.

So now that we know whether it opens upward or downward, then we know where the region between the zero's is positive  or negative and if the region outside the zero's is positive or negative. So we now know the region where the velocity is positve or negative and have found the answer to the problem.

Is this cheating? Is there a more elegent way I could have solved this? A way to find the points on a parabolla that are above or below the x axis without considering its graph and whether the parabolla opens upward or downward?

The graph of the velocity in this problem was a parabolla which is pretty easy since it only has one vertice. But an equation of higher degree would be a nightmare! Is there any simple method to determine all the points above or below the graph of a function? You can't always say "f(x) > 0" or "f(x) < 0" and solve, because many functions have more then one zero and some of the roots will not be consitant with the inequality.

Another strange problem I got. Find the postive number that exceeds its cube by the greatest amount.

At first I tried differentiating c = x^3

3x^2

then setting it equal to zero:

3x^2 = 0

Solved, x = 0. Thats obviously incorrect. I was unsure how to solve the problem and tried:

x > x^3

differentiated:

1 > 3x^2

1/3 > x^2        divided both sides by 3

sqrt(1/3) > x    square root of both sides

sqrt(3)/3 > x   rationalized denominator

This expression is an inequalit, but the answer to the problem was x equals sqrt(3)/3.

Peculiar. If I had set x = x^3, differentiated, and solved. That would have given me the answer.

But why? It doesn't seem like a normal maximums and minimums problem.