Use the Binomial Theorem to find the numerical value of (1.001)^5 correct to five decimal places

Use the Binomial Theorem to find the numerical value of...

Key Concepts

Rounding and Precision

Rounding involves adjusting a numerical value to a specified degree of precision, such as to a certain number of decimal places. This concept ensures that the result of a numerical approximation, like a series expansion, is reported in a clear and concise manner. Rounding is particularly important when the exact value is either irrational or overly complex, allowing for simpler, yet appropriately accurate, representations.

Binomial Theorem

The Binomial Theorem provides a method to expand expressions raised to a power, formulated as (a + b)^n = ? (n choose k) * a^(n-k) * b^k, where the summation is from k = 0 to n. This theorem is fundamental in algebra and combinatorics because it links combinatorial concepts with polynomial expressions, allowing for systematic expansion without directly multiplying the expression multiple times.

Numerical Approximation

Numerical approximation is the process of finding approximate values using computational techniques. When exact arithmetic is cumbersome or unnecessary, approximations provide a practical solution, particularly in cases involving series expansion like the Binomial Theorem. It is crucial for evaluating expressions to a certain number of decimal places, ensuring that results are both manageable and sufficiently accurate for practical applications.

Transcript

00:01 If we wanted to find the value of 1 .001 to the 5th, we can use a calculator, or we could use the binomial theorem written above in black.

00:14 So 1 .001 to the 5th can be rewritten as 1 plus 10 to the negative 3rd all to the 5th.

00:24 Looking at this equation, we could see that in this form, x equals 1, a equals 10 to the 3rd, all to the 5th.

00:31 The negative third and n equals 5.

00:36 We can expand this using the binomial theorem as shown above.

00:42 So for the first term, j would equal 0.

00:46 So we get n choose j, or 5 choose 0, times x, which in our case is 1, to the n minus j.

00:58 So to the 5 minus 0 power times a, which is 10 to the negative 3 to the j, the j.

01:06 Which for this term again is zero the next term would be 5 choose 1 times 1 to the 5 minus 4 times 10 to the negative 3rd to the first for the third term j equals 2 so we get 5 choose 2 1 to the 5 minus 2 sorry for this term it should be 5 minus 1 i don't know why i wrote four there.

01:41 So going back to this term, 5 choose 2 times 1 to the 5 minus 2 times 10 to the negative 3 to the second power.

01:52 For the fourth term, j equals 3.

01:55 So we get 5 choose 3 times 1 to the 5 minus 3 times 10 to the negative 3 to the 3 to the 3.

02:09 Now, for the fifth term, j is equal to 4.

02:14 So we get 5 choose 4, 1 to the 5 minus 4 power, times 10 to the negative 3 to the 4th.

02:26 And for the last term, j equals 5.

02:28 So we get 5 choose 5 times 1 to the 5 minus 5 times 10 to the negative 3 to the 5 to the 5.

02:39 So we could simplify this a lot.

02:45 We can use pascal's triangle to find the terms 5 to 0 all the way up to 5 choose 5.

02:54 Because this is a fifth degree binomial, we would use the fifth row of pascal's triangle.

03:02 So 520 equals 1.

03:06 1 to any power equals 1.

03:08 And 10 to the negative third to the 0 also equals 1.

03:14 So each of these terms equals 1.

03:17 So the first term is all equal to 1...

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