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    Hey I'm doing past questions and I've drawn a blank..
    Part one of the Q is to expand (1+3x)^8 up to the term in x^3 (which I've done).
    Part two is to use this series to estimate the value of (1.003)^8, stating the value of x that I use. How do I estimate this up to the power of 8 when I only have the series up to the term of x^3? I gather I'm not meant to just expand the whole thing since it's an estimation question. Thanks!
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    (Original post by mueslii)
    Hey I'm doing past questions and I've drawn a blank..
    Part one of the Q is to expand (1+3x)^8 up to the term in x^3 (which I've done).
    Part two is to use this series to estimate the value of (1.003)^8, stating the value of x that I use. How do I estimate this up to the power of 8 when I only have the series up to the term of x^3? I gather I'm not meant to just expand the whole thing since it's an estimation question. Thanks!
    That's not quite what it means. You are saying that (1+3x)^8 \approx a+bx+cx^2+dx^3 (for whatever a,b,c,d you got in your answer). Now this is merely an approximation. If you were to include higher and higher powers of x, you would get a more accurate approximation of the true value.

    So when the question asks you to estimate some number to the nth power,.it does NOT mean that you NEED your sum to go up to the nth power of x. You get the approximation with however many powers of x you included in your expansion, so more means more accurate as per my comment above, but since you expanded up to the cube term, you just need to use that expansion.

    All you need to do is choose an appropriate x such that you get (1.003)^8 when you plug it into your expression.
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    (Original post by RDKGames)
    That's not quite what it means. You are saying that (1+3x)^8 \approx a+bx+cx^2+dx^3 (for whatever a,b,c,d you got in your answer). Now this is merely an approximation. If you were to include higher and higher powers of x, you would get a more accurate approximation of the true value.

    So when the question asks you to estimate some number to the nth power,.it does NOT mean that you NEED your sum to go up to the nth power of x. You get the approximation with however many powers of x you included in your expansion, so more means more accurate as per my comment above, but since you expanded up to the cube term, you just need to use that expansion.

    All you need to do is choose an appropriate x such that you get (1.003)^8 when you plug it into your expression.
    Ohh I see now, I think misunderstanding the question is why I was so confused. Thank you!
 
 
 
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