bigmansouf
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Question:
Find the first four terms of the expansion of  (1-8x)^{0.5} in ascending of x. Substitute  x = \frac{1}{100} and obtain the value of  \sqrt{23} correct to five significant figures

a) find the first four terms of the expansion of  (1-8x)^{0.5} in ascending of x.
 (1-8x)^{0.5} = \left ( 1 + ((0.5)(-8x))+\frac{(0.5)(-0.5)(-8x)^{2}}{1 \times 2} + \frac{(0.5)(-0.5)(-1.5)(-8x)^{3}}{1 \times 2 \times 3} \right )
 \left ( 1-4x -8x^2 - 32x^3 \right )

b)Substitute  x = \rac{1}{100} and obtain the value of  \sqrt{23} correct to five significant figures
I have problem with this part

now it says to sub  x = \frac{1}{100}
so;  \left ( 1 - (8\frac{1}{100}) \right )^{0.5} = 0.9591663047

using the expansion;

 \left ( 1-4x -8x^2 - 32x^3 \right )

i sub x = 1/00 into the expansion

 \left ( 1-\left (4 \left(\frac{1}{100}\right) \right) -\left (8\left(\frac{1}{100}\right)^{2} \right) - \left ( 32\left ( \frac{1}{100} \right )^{3} \right ) \right )

it produces 0.959168
i know that  \sqrt{23} =4.795831523

My issue is that I dont know why I am not getting 4.795831523
since part b is a simple substitution question asking me to sub x = 1/100 shouldn't the I get the answer 4.795831523, when i sub x= 1/100 into the expansion or binomial.

thank you for helping me
Last edited by bigmansouf; 1 year ago
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mqb2766
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X = 1/100 does not produce sqrt(23) for
(1 - 8x)^0.5
Just sub in the value?

How can you transform
(1 - 8/100)^0.5
Into sqrt(23)?
(Original post by bigmansouf)
Question:
Find the first four terms of the expansion of  (1-8x)^{0.5} in ascending of x. Substitute  x = \frac{1}{100} and obtain the value of  \sqrt{23} correct to five significant figures

a) find the first four terms of the expansion of  (1-8x)^{0.5} in ascending of x.
 (1-8x)^{0.5} = \left ( 1 + ((0.5)(-8x))+\frac{(0.5)(-0.5)(-8x)^{2}}{1 \times 2} + \frac{(0.5)(-0.5)(-1.5)(-8x)^{3}}{1 \times 2 \times 3} \right )
 \left ( 1-4x -8x^2 - 32x^3 \right )

b)Substitute  x = \rac{1}{100} and obtain the value of  \sqrt{23} correct to five significant figures
I have problem with this part

now it says to sub  x = \frac{1}{100}
so;  \left ( 1 - (8\frac{1}{100}) \right )^{0.5} = 0.9591663047

using the expansion;

 \left ( 1-4x -8x^2 - 32x^3 \right )

i sub x = 1/00 into the expansion

 \left ( 1-\left (4 \left(\frac{1}{100}\right) \right) -\left (8\left(\frac{1}{100}\right)^{2} \right) - \left ( 32\left ( \frac{1}{100} \right )^{3} \right ) \right )

it produces 0.959168
i know that  \sqrt{23} =4.795831523

My issue is that I dont know why I am not getting 4.795831523
since part b is a simple substitution question asking me to sub x = 1/100 shouldn't the I get the answer 4.795831523, when i sub x= 1/100 into the expansion or binomial.

thank you for helping me
Last edited by mqb2766; 1 year ago
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Pangol
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(Original post by bigmansouf)
Question:
Find the first four terms of the expansion of  (1-8x)^{0.5} in ascending of x. Substitute  x = \frac{1}{100} and obtain the value of  \sqrt{23} correct to five significant figures

a) find the first four terms of the expansion of  (1-8x)^{0.5} in ascending of x.
 (1-8x)^{0.5} = \left ( 1 + ((0.5)(-8x))+\frac{(0.5)(-0.5)(-8x)^{2}}{1 \times 2} + \frac{(0.5)(-0.5)(-1.5)(-8x)^{3}}{1 \times 2 \times 3} \right )
 \left ( 1-4x -8x^2 - 32x^3 \right )

b)Substitute  x = \rac{1}{100} and obtain the value of  \sqrt{23} correct to five significant figures
I have problem with this part

now it says to sub  x = \frac{1}{100}
so;  \left ( 1 - (8\frac{1}{100}) \right )^{0.5} = 0.9591663047

using the expansion;

 \left ( 1-4x -8x^2 - 32x^3 \right )

i sub x = 1/00 into the expansion

 \left ( 1-\left (4 \left(\frac{1}{100}\right) \right) -\left (8\left(\frac{1}{100}\right)^{2} \right) - \left ( 32\left ( \frac{1}{100} \right )^{3} \right ) \right )

it produces 0.959168
i know that  \sqrt{23} =4.795831523

My issue is that I dont know why I am not getting 4.795831523
since part b is a simple substitution question asking me to sub x = 1/100 shouldn't the I get the answer 4.795831523, when i sub x= 1/100 into the expansion or binomial.

thank you for helping me
You can only have worked out the value of the LHS when x = 1/100 with a calculator. This defeats the object of the question - the idea of binomial approximations in this context is to find the approximate value of roots by using only addition and multiplication of rational numbers.

When you substitute x = 1/100 into the LHS, you get (1 - 8/100)^0.5 = (1 - 2/25)^0.5 = (23/25)^0.5. Does that give you any ideas?
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bigmansouf
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#4
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(Original post by mqb2766)
X = 1/100 does not produce sqrt(23) for
(1 - 8x)^0.5
Just sub in the value?

Can you transform
(1 - 8/100)^0.5
Into sqrt(23)?
yes i can i get  \left ( \frac{23}{25} \right )^{0.5}
i see now i thought that binomial theorem can be used to find approx but this question is not asking me the approximate to the square roof of 23
i thought that it was an approximate type of question. I personally do not see the point of part b and what it is trying to teach me
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Pangol
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(Original post by bigmansouf)
yes i can i get  \left ( \frac{23}{25} \right )^{0.5}
i see now i thought that binomial theorem can be used to find approx but this question is not asking me the approximate to the square roof of 23
i thought that it was an approximate type of question. I personally do not see the point of part b and what it is trying to teach me
It is asking for an approximation to root 23. Keep going. Split up the (23/25)^0.5.
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mqb2766
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So multiply the answer you got by 5.
(Original post by bigmansouf)
yes i can i get  \left ( \frac{23}{25} \right )^{0.5}
i see now i thought that binomial theorem can be used to find approx but this question is not asking me the approximate to the square roof of 23
i thought that it was an approximate type of question. I personally do not see the point of part b and what it is trying to teach me
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bigmansouf
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(Original post by Pangol)
You can only have worked out the value of the LHS when x = 1/100 with a calculator. This defeats the object of the question - the idea of binomial approximations in this context is to find the approximate value of roots by using only addition and multiplication of rational numbers.

When you substitute x = 1/100 into the LHS, you get (1 - 8/100)^0.5 = (1 - 2/25)^0.5 = (23/25)^0.5. Does that give you any ideas?
 (23/25)^0.5 can be changed to   \left ( 1-\left (4 \left(\frac{1}{100}\right) \right) -\left (8\left(\frac{1}{100}\right)^{2} \right) - \left ( 32\left ( \frac{1}{100} \right )^{3} \right ) \right )

then to transfer to  \sqrt{23} by multiplying   \left ( 1-\left (4 \left(\frac{1}{100}\right) \right) -\left (8\left(\frac{1}{100}\right)^{2} \right) - \left ( 32\left ( \frac{1}{100} \right )^{3} \right ) \right ) by 5

thus;

 5 \times  \left ( 1-\left (4 \left(\frac{1}{100}\right) \right) -\left (8\left(\frac{1}{100}\right)^{2} \right) - \left ( 32\left ( \frac{1}{100} \right )^{3} \right ) \right )
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bigmansouf
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i get it now
thanks
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Pangol
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(Original post by bigmansouf)
 (23/25)^0.5 can be changed to   \left ( 1-\left (4 \left(\frac{1}{100}\right) \right) -\left (8\left(\frac{1}{100}\right)^{2} \right) - \left ( 32\left ( \frac{1}{100} \right )^{3} \right ) \right )

then to transfer to  \sqrt{23} by multiplying   \left ( 1-\left (4 \left(\frac{1}{100}\right) \right) -\left (8\left(\frac{1}{100}\right)^{2} \right) - \left ( 32\left ( \frac{1}{100} \right )^{3} \right ) \right ) by 5

thus;

 5 \times  \left ( 1-\left (4 \left(\frac{1}{100}\right) \right) -\left (8\left(\frac{1}{100}\right)^{2} \right) - \left ( 32\left ( \frac{1}{100} \right )^{3} \right ) \right )
That's it - so it is an approximation question!
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mqb2766
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One thing its trying to teach you is that while the expansion is only valid for
|x|<1/8
So for sqrt of 0 to 2 .
You can use transformations to apply it to other numbers outside this range.
(Original post by bigmansouf)
yes i can i get  \left ( \frac{23}{25} \right )^{0.5}
i see now i thought that binomial theorem can be used to find approx but this question is not asking me the approximate to the square roof of 23
i thought that it was an approximate type of question. I personally do not see the point of part b and what it is trying to teach me
Last edited by mqb2766; 1 year ago
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