Section+3.4

This page is still a work in progress. We started out class by doing Check You Understanding problem 1 in Section 3.4 to make sure we were all on the same page about synthetic division and taylor expansions. Below is problem 1 and what you would do to solve it.

1. Suppose that media type="custom" key="11548464" and that media type="custom" key="11548468". Expand the following in powers of media type="custom" key="11548478".

a.media type="custom" key="11548506" b.media type="custom" key="11548512" c.media type="custom" key="11548522" d.media type="custom" key="11548526" e. media type="custom" key="11548530"

a. For part a we can use synthetic division to find the coefficients before each term in the taylor expansion by putting a 3 in the box. Remember that the number in the box is always the opposite of the number in the factor that you are trying to expand at. For example, if you are trying to expand at (x-3), you would put a three in the box. if you were trying to expand at media type="custom" key="11548454", you would put a -2 in the box. Mrs. Tyson encouraged us to put a box around the last number we find for every time you divide by the number in the box. This is really helpful at the end because we can see which numbers we will be using as coefficients. Also, the number of boxes you have once you've completed all the division will be the degree of the taylor expansion you end up with.



(Mrs. Tyson doesn't have a color scanner. The author had the synthetic division color coded to match with the equation below)

As you can see there are three boxes in the above synthetic division, so that will be the degree of the taylor expansion.

You end up with a taylor expansion of: media type="custom" key="11548538"

b. You can also use synthetic division for this problem because it is essentially the same as part a except with different number.



you end up with the taylor expansion of: media type="custom" key="11548554"

c. For this part you could addmedia type="custom" key="11548594" and media type="custom" key="11548612"together and then do synthetic division with the resulting function, or you could simply add their taylor expansions together by adding like terms together: media type="custom" key="11548562" media type="custom" key="11548568"

d. For this part, like part c, you could multiplymedia type="custom" key="11548586" by four then do all that synthetic division all over again, or you could just multiply its taylor expansion by 4 by simply multiplying each coefficient by 4. media type="custom" key="11548576" media type="custom" key="11548622"

e. Again, on this part you could multiply f(x) by 4, then subtract g(x), then do the synthetic division with your new function, or you could use your taylor expanded answer from part d, then simply subtract the taylor expansion of g(x) from it, which you have also already done in part b, by subtracting like terms. (Like terms are in the same color.) media type="custom" key="11548632" media type="custom" key="11548634"

Mrs. Tyson then told us that we can do synthetic division by putting a variable in the box, so we get general answers for a specific function. Later, we found out that this is really useful because if you are dealing with one function, but you want to find information about it for a number of different inputs (like the tangent line or slope of the tangent line), doing the synthetic division with a arbitrary variable shortens the process by a lot. The example Mrs. Tyson gave in class is the following:

Expandmedia type="custom" key="11548638"



The taylor expansion you will end up with is: media type="custom" key="11548646"

Finally, we looked at the dialogue between Derman and Tony in section 3.4 and we discovered that when you divide any polynomial by a power of x, the remainder is going to be every term in the polynomial that has a power of x that is less than the power of x you divided by. For example, let's work with the polynomial media type="custom" key="11548650" and divide by the following powers of x:

1. x. When you divide this polynomial by x, the remainder will be 31 2. media type="custom" key="11548658". When you divide this polynomial by media type="custom" key="11548668", the remainder will be media type="custom" key="11548670". 3. media type="custom" key="11548682". When you divide this polynomial by media type="custom" key="11548684" the remainder will bemedia type="custom" key="11548690".