Difference between revisions of "Mathematical Methods in the Physical Sciences/Chapter 1/Section 2"
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The <math>\sum</math> symbol may be new. Well, it's no longer new after this chapter. Get used to it. It is going to be your best friend for a long time. | The <math>\sum</math> symbol may be new. Well, it's no longer new after this chapter. Get used to it. It is going to be your best friend for a long time. | ||
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| + | == Problems == | ||
| + | |||
| + | === 1 === | ||
| + | === 2 === | ||
| + | |||
| + | The <math>(-1)^n</math> is the secret sauce for how to get alternating terms. Be careful: if the 1st term is positive, then you need <math>-1(-1)^n</math> or <math>(-1)^{n+1}</math>. | ||
| + | |||
| + | === 3 === | ||
| + | === 4 === | ||
| + | === 5 === | ||
| + | === 6 === | ||
| + | === 7 === | ||
| + | === 8 === | ||
| + | |||
| + | When you have consecutive multiplied numbers, then factorials could be involved. | ||
| + | |||
| + | Note that <math>5!/3!</math> is simply <math>5 \cdot 4</math>. The 3, 2, 1 terms disappear by dividing by <math>3!</math>. | ||
| + | |||
| + | Of course, you can always do <math>(n+1)(n)(n-1)</math> or something similar, but I think factorials are cooler. | ||
| + | |||
| + | === 9 === | ||
| + | === 10 === | ||
| + | === 11 === | ||
| + | === 12 === | ||
[[../Section 3|Next Section]] | [[../Section 3|Next Section]] | ||
Latest revision as of 10:28, 19 April 2012
Contents
Overview
This is a notation chapter. You're simply becoming more familiar with the terminology and notation of infinite series.
The <math>\sum</math> symbol may be new. Well, it's no longer new after this chapter. Get used to it. It is going to be your best friend for a long time.
Problems
1
2
The <math>(-1)^n</math> is the secret sauce for how to get alternating terms. Be careful: if the 1st term is positive, then you need <math>-1(-1)^n</math> or <math>(-1)^{n+1}</math>.
3
4
5
6
7
8
When you have consecutive multiplied numbers, then factorials could be involved.
Note that <math>5!/3!</math> is simply <math>5 \cdot 4</math>. The 3, 2, 1 terms disappear by dividing by <math>3!</math>.
Of course, you can always do <math>(n+1)(n)(n-1)</math> or something similar, but I think factorials are cooler.
