C4 binomial expansion

Hi,

I have two questions, both about the values for which binomial series are valid.

The first question is, what determines, whether or not we use a $\leq$ sign or just a < sign? The answers in my book vary with which one they use, and there doesn't appear to be any discernible difference in the question.

My second question is, when we have to take a factor out of the parenthesis, for example $(16 + y)^{\frac{1}{2}}$ when we want to find the value for which the series is valid, do we put the 16 back in? Again, in my book, sometimes they do and sometimes they don't, with no explanation as to why.

Any help would be really appreciated!
Thanks
Marc.

Hi,

I have two questions, both about the values for which binomial series are valid.

The first question is, what determines, whether or not we use a $\leq$ sign or just a < sign? The answers in my book vary with which one they use, and there doesn't appear to be any discernible difference in the question.

My second question is, when we have to take a factor out of the parenthesis, for example $(16 + y)^{\frac{1}{2}}$ when we want to find the value for which the series is valid, do we put the 16 back in? Again, in my book, sometimes they do and sometimes they don't, with no explanation as to why.

Any help would be really appreciated!
Thanks
Marc.

<
putting in or keeping out gives the sam answer

just remember that when you take out the 16 you must apply the power 1/2 to it

The first question is, what determines, whether or not we use a $\leq$ sign or just a < sign? The answers in my book vary with which one they use, and there doesn't appear to be any discernible difference in the question.

<
putting in or keeping out gives the sam answer

Hi, thanks for your help. Yeah, I remember that the 16 also has the same power, so it would be 4.

But in my book, for $(4+2x)^{\frac{1}{2}} = 2(1+\frac{1}{2}x)$ it says it is only valid $x \leq 2$



But for $(16+y)^{\frac{1}{2}} =4(1+\frac{y}{16})$ the book says it is valid when the mod of y is less than 4. So how did they work this out?

it sounds like the book is wrong... if you have ( 1 + y/16) in the bracket then y is between -16 and 16

The interval of validity does need to have less-than, not less-than-or-equal to. Consider $(1+x)^-1 = 1-x+x^2-x^3+\dots$. Substitute -1 for x, and you get $1+1+1+\dots$ which obviously doesn't converge - and yet the convergence interval for this series is $(-1,1)$. For each question, you need to check whether the sum converges at the endpoints of the interval. (Actually, it seems plausible that it never converges on the endpoints - not sure about that without actually putting pencil to paper.)

OK, this makes sense to me, thanks

So simply, to get the value for which it is valid, we just look inside the brackets, and ignore any factors that we've taken out?

yes.

So, in all cases, should we say 'less than', rather than 'equal to or less than'?

Hi, thanks for your help. Yeah, I remember that the 16 also has the same power, so it would be 4.

But in my book, for $(4+2x)^{\frac{1}{2}} = 2(1+\frac{1}{2}x)^{\frac{1}{2}}$ it says it is only valid $x \leq 2$



But for $(16+y)^{\frac{1}{2}} =4(1+\frac{y}{16})$ the book says it is valid when the mod of y is less than 4. So how did they work this out?

Your book seems odd

< as I said, always

and your |y|<16 either way

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