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im working on some basic calculus homework (applying the Chain Rule, Power rule, Quotient rule, etc).
i guess my question could be solved by anyone who knows math--not calculus, as it just involves simplifying.
after completing a question and comparing to the answer in the book, they appear different. when i substitute a value (tested for x=1 and x=2) and evaluate my formula and the book's answer's formula, the same answer is given.. which leads me to believe i have the same answer--just longer.
so my question is can someone show me how to simplify (my answer) the following:
Quote:
(2x + 1)(5(4x - 1)^4 * 4) + [(4x - 1)^5 * 2]
to (the books answer):
Quote:
6(8x + 3)(4x - 1)^4
i realize how bad it looks in a simple text editor, but sorry.
Distribution: Ubuntu, Debian, Various using VMWare
Posts: 2,088
Rep:
Im not going to do it for you, because you learn better by doing
Basically, if you just expand you answer out, you might be able to see some factorisation that you can do. You can take (4x-1)^4 as a factor, and then try expanding and simplifying the other factor.
OK, I had figured it out.. Thanks for the input IBall. brianL, thanks for your interest.. i had arrived at an anwser that was the same as the books, just not as simplified. does Maxima show step-by-step simplifications of formulas? i didnt try it so i dont know, but i assumed it didnt.
OK, I had figured it out.. Thanks for the input IBall. brianL, thanks for your interest.. i had arrived at an anwser that was the same as the books, just not as simplified. does Maxima show step-by-step simplifications of formulas? i didnt try it so i dont know, but i assumed it didnt.
Unless you are studying independently, why not go track down your professor for such problems? I have been in too many math classes where other people were too afraid to ask questions...the end result is the professor has no idea how good of a job he/she is doing at teaching and so he/she ends up going too fast for the class. Also, coming in to visit demonstrates how serious you are about learning, so when it comes time for final grades and your grade is tottering between an A and A-, they will remember your effort...
In a modern physics class, I made the mistake of asking the physics professor to explain how he got from one step to the next. I was following along and that step lost me. He spent the rest of the class and went through 3 blackboards full of equations. I doubt any one else was able to follow that step either, but the rest of the class either though it was very funny, or was mad at me because they weren't interested.
It's been so long since I was in school, I forgot far more math than I remember.
I felt so proud of myself in high school, when I worked out the formula for the sum of a series, on my own. Only to find out later that Euler did it when he was 8 years old. By 9 RMS was learning calculus. WOW!
Unless you are studying independently, why not go track down your professor for such problems?
this week is our 'reading week' and he told us this week his office hours wouldnt be very reliable. we have a class forum which i posted the same question on, but no one replied. but after i figured it out i posted on that forum that it had been solved.
regarding the marking scheme, i go to a very populated university with afew hundred in my math class and i doubt me going in to see him would change his final marking scheme against me. when i was in college, which was much smaller, i did this alot though.. and it does seem to be true.
jschiwal, figuring that formula out on your own is quite impressive, i would say.
Not really. It was very easy. I was looking at a series like this:
1 2 3 4 5 6 7 8 9 10
and noticed this pattern:
Code:
1 2 3 4 5
10 9 8 7 6
Notice how the sums are the same. There are n/2 columns which each add up to (n+1). So the sum of the series is n(n+1)/2. I told you it was easy. I then tested the formula for when n is odd. Although you could also start the series with 0 in that case.
Now for fun you can use the same method to determine the sum of an integer series m ... n where n > m.
Here is a fun one you can show your friends. Explain why e to the minus pi power is equal to -1.
---
The answer is that e to the -x power describes a point traveling a unit circle, where x is the angle traveled in radians. Since pi radians is a half circle, you know that the point rests on the minus x axes. With just that picture in mind, you can pop off the answer to e to the minus pi/4.
To really amaze your friends give a bunch of answers first before the explanation.
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