I understand the method of finding the range of validity however I am confused on what its significance is, I don't understand what it means, and why is there a modulus? can you explain it to me please?
the range of validity tells you the values of x for which the expansion is valid. For infinite series expansions, they don't work for all values of x. So for example, (1-x)^(-1) = 1 + x + x^2 + x^3 + ... for |x| < 1. This means the expansion only works for -1 < x < 1. So if we sub in 2: LHS: (1-2)^(-1) = -1 RHS: 1 + 2 + 2^2 + 2^3 + ... which is definitely not equal to -1 If we sub in 0.5 LHS: (1-0.5)^(-1) = 2 RHS: 1 + 0.5^2 + 0.5^3 + ... which is equal to 2. So the expansion worked for a number between -1 and 1, but not for one outside this range of validity. These are just two examples, which certainly isn't a proof, but it shows what we're talking about. The modulus version of the range of validity |x| < a means exactly the same thing as -a < x < a
I don't know if this is a correct question to ask but, why is it that for the (1 + x) ^ n expansion, the range of validity if |X| < 1? Why the modulus?
Hi. How do we find valid value of x for | 1/(3x) | < 1 ? It's confusing because x is apart of the denominator.... So how do I write it in -1< x < 1 form ?
First sketch y = |1/(3x)|. Draw a horizontal line at y = 1. www.desmos.com/calculator/z7wwmcqxpe Then solve |1/(3x)| = 1: 1/(3x) = 1 ⇒ 3x = 1 ⇒ x = 1/3 -1/(3x) = 1 ⇒ 3x = -1 ⇒ x = -1/3 Then answer the question "Where is the graph of y= |1/(3x)| BELOW y = 1?". The answer is (∞,-1/3)∪(1/3,∞)
what if im given a question involving both binomial and partial fraction. I have to expand 2 times and then add. but now its asking me how to write down the set of values for which expression is valid. in this case it was ((1-x)^-1) + ((1+x/2)^-1) - ((1+x/2)^-2) after expressing in partial fractions. how to do it???? help plz
well, I think first you would expand the brackets to get (2-7x)^n, then factorising 2 out to get 2^n(1-7x/2) and repeat the process shown in the video: |-7x|
Instead of bringing the 2^1/3 outside of the brackets (which is slightly more complicated), couldn't you just do |-7x| < 2 and then divide through by 7 (mod takes care of the neg.) giving you |x| < 2/7
Elliot 73 This is true. However, in general, if you were required to expand (2-7x)^(1/3), you will need to factor out the 2^(1/3) in order to use the formula (1+x)^n as seen in the formula booklet. The likelihood would be that a question would ask you to expand first and then ask about which values it would be valid for afterwards, so the factoring out would already have occurred.
No matter how many times I learn this I always seem to forget it
Same! I keep coming back to this video
Legend! Excellent video, explained very simply but gets the point across perfectly!
You must be a very good teacher/lecturer :)
Fantastic video!
very clear , thanks Jack
What if there is a composite function for example ln(1+x) + cosx, would the range be -1
Woah..This video was really helpful.
Thanks a lot.
Very well explained. Thanks
Thanks a bunch! You rock!
I understand the method of finding the range of validity however I am confused on what its significance is, I don't understand what it means, and why is there a modulus? can you explain it to me please?
the range of validity tells you the values of x for which the expansion is valid. For infinite series expansions, they don't work for all values of x. So for example, (1-x)^(-1) = 1 + x + x^2 + x^3 + ... for |x| < 1. This means the expansion only works for -1 < x < 1.
So if we sub in 2:
LHS: (1-2)^(-1) = -1
RHS: 1 + 2 + 2^2 + 2^3 + ... which is definitely not equal to -1
If we sub in 0.5
LHS: (1-0.5)^(-1) = 2
RHS: 1 + 0.5^2 + 0.5^3 + ... which is equal to 2.
So the expansion worked for a number between -1 and 1, but not for one outside this range of validity. These are just two examples, which certainly isn't a proof, but it shows what we're talking about.
The modulus version of the range of validity |x| < a means exactly the same thing as -a < x < a
Jack Brown thank you so much!
Why is there not a negative sign when shifting the fraction to the right side during finding validity?
The minus sign is "absorbed" by the modulus in the way that |-3| = |3|
I don't know if this is a correct question to ask but, why is it that for the (1 + x) ^ n expansion, the range of validity if |X| < 1? Why the modulus?
|x|
TLMaths Am I wrong to say this is similar to Sum to Infinity in a Geometric Progression?
You're not wrong
Hi. How do we find valid value of x for
| 1/(3x) | < 1 ? It's confusing because x is apart of the denominator.... So how do I write it in -1< x < 1 form ?
First sketch y = |1/(3x)|.
Draw a horizontal line at y = 1.
www.desmos.com/calculator/z7wwmcqxpe
Then solve |1/(3x)| = 1:
1/(3x) = 1 ⇒ 3x = 1 ⇒ x = 1/3
-1/(3x) = 1 ⇒ 3x = -1 ⇒ x = -1/3
Then answer the question "Where is the graph of y= |1/(3x)| BELOW y = 1?". The answer is (∞,-1/3)∪(1/3,∞)
what if im given a question involving both binomial and partial fraction. I have to expand 2 times and then add. but now its asking me how to write down the set of values for which expression is valid.
in this case it was ((1-x)^-1) + ((1+x/2)^-1) - ((1+x/2)^-2) after expressing in partial fractions.
how to do it???? help plz
Well the first expansion is valid for |x|
Thanks a lot sir. I'm having a test tomorrow
How do you know whether to use the ≤ or < symbol?, in a question I did (4+2x)^1/2 the answer in the book stated lxl ≤ 2
+Alec09039 It must be a typo, it should always be
Jack Brown Thanks
Nice explanation
Absolute legend
In the last question if you take (1+(1-7x))^ now apply the condition on mod |1-7x|
I think this is a bit beyond what I know as really we're asking that if (1+x)^n is valid for |x|
well, I think first you would expand the brackets to get (2-7x)^n, then factorising 2 out to get 2^n(1-7x/2) and repeat the process shown in the video: |-7x|
what if you have a quadratic function let say power half maybe, how will you find the range..or how will you be able to 1 plus or minus something..
Can you give me an actual example of what you mean?
@@TLMaths square root of (x^2-2+x)^7
So with this particular example you've chosen, you really start to open up a can of worms. Firstly, x^2 + x - 2 is negative for -2
@@TLMaths thank you sir
What if the questions ask you to expand sin^-1x and state the range of values of x for which the expansion is valid
It won't, it'll only ask for a binomial expansion.
Good video
life saver, thanks for this!
Simple, easy, perfect. Thanks ^_^
Bravo! thanks.
God amongst men 🙏
Instead of bringing the 2^1/3 outside of the brackets (which is slightly more complicated), couldn't you just do |-7x| < 2 and then divide through by 7 (mod takes care of the neg.) giving you |x| < 2/7
Elliot 73 This is true. However, in general, if you were required to expand (2-7x)^(1/3), you will need to factor out the 2^(1/3) in order to use the formula (1+x)^n as seen in the formula booklet. The likelihood would be that a question would ask you to expand first and then ask about which values it would be valid for afterwards, so the factoring out would already have occurred.
Jack Brown Ah, yeah, that makes sense. Thanks for the reply!
Thanks a lot!
Thankss!!!
What if it was (1+x^2)^2?
Do you mean if it was (1+x^2)^(-2) ?
@@TLMaths yes
Then replace x with x^2 in the range of validity:
|x^2| < 1
so -1 < x < 1
@@TLMaths thanks!!
bruh amazing!!!!!!!!!!!!!!
🐐
thanks
My teacher literally copied his entire lecture from this video with every SAME example and explained it in the class. 😑
If it helps, that can only be a good thing. Remember, your teacher is learning too as it may be their first time teaching this topic.
thankyou.