Help with Edexcel Further Pure challenge question

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Hi, this question is the challenge question from Edexcel Further Pure 1, Exercise 7A. The steps in the solution bank aren't fully making sense, just wondering if anyone who knows how to do it could answer it again with some explanations? The main issue is the jump from the second step with an equals sign to the step below it. Thank you.

Reply 1

mqb2766

21

Original post

by loik24

Hi, this question is the challenge question from Edexcel Further Pure 1, Exercise 7A. The steps in the solution bank aren't fully making sense, just wondering if anyone who knows how to do it could answer it again with some explanations? The main issue is the jump from the second step with an equals sign to the step below it. Thank you.

Can you be clearer about which part youre asking about?

Reply 2

loik24

OP

12

Original post

by mqb2766

Can you be clearer about which part youre asking about?

The jump from the second step starting with an equals sign to the one below it please

Reply 3

mqb2766

21

Original post

by loik24

The jump from the second step starting with an equals sign to the one below it please

So going from the third line to the fourth line where they switch the summation and differentiation?

Reply 4

loik24

OP

12

Original post

by mqb2766

So going from the third line to the fourth line where they switch the summation and differentiation?

Sorry, the line after they use product rule and we see (n k-1). I don’t understand what they did to go between those lines

Reply 5

mqb2766

21

Original post

by loik24

Sorry, the line after they use product rule and we see (n k-1). I don’t understand what they did to go between those lines

The superscripts represent the number of times the functions have been differentiated, so they just apply the product rule and increase the appropriate numbers by 1, so
f^k -> f^(k+1)
g^(n-k) -> g^(n-k+1)

Reply 6

loik24

OP

12

Original post

by mqb2766

So going from the third line to the fourth line where they switch the summation and differentiation?

Sorry still don’t quite get it? What I’m really asking is that the first term in both lines are the same - what do they do and why is the second term transformed to what it is? Like why have they removed one from f(x)’s superscript and added one to g(x)? And then why has the limits of the summation changed, as well as now n choose k-1 compared to when it was n choose k?

Reply 7

mqb2766

21

Original post

by loik24

Sorry still don’t quite get it? What I’m really asking is that the first term in both lines are the same - what do they do and why is the second term transformed to what it is? Like why have they removed one from f(x)’s superscript and added one to g(x)? And then why has the limits of the summation changed, as well as now n choose k-1 compared to when it was n choose k?

They just done the product rule / differentiated on
nCk f^(k)(x) g^(n-k)(x)
nCk is a constant multiplier so its unchanged and f^(k) and g^(n-k) are two functions which are multiplied together. So to differentiate it using the product rule
fg' + f'g
and when differentiated g^(n-k) -> g^(n-k+1) as youve differentiated it one more time and f^(k) -> f^(k+1) as youve differentiate it one more time.

On the next line, theyve modified the summation limits and superscripts of f^(k+1) g^(n-k), but the overaell summation stays the same which is easy enough to verify by subbing the new k=1 to n+1 in and noting it refers to the same series as the old k=0 to n. The problem solving box in the question covers this part.

Reply 8

loik24

OP

12

Original post

by mqb2766

They just done the product rule / differentiated on
nCk f^(k)(x) g^(n-k)(x)
nCk is a constant multiplier so its unchanged and f^(k) and g^(n-k) are two functions which are multiplied together. So to differentiate it using the product rule
fg' + f'g
and when differentiated g^(n-k) -> g^(n-k+1) as youve differentiated it one more time and f^(k) -> f^(k+1) as youve differentiate it one more time.
On the next line, theyve modified the summation limits and superscripts of f^(k+1) g^(n-k), but the overaell summation stays the same which is easy enough to verify by subbing the new k=1 to n+1 in and noting it refers to the same series as the old k=0 to n. The problem solving box in the question covers this part.

Ohhh thank you, that’s okay. Then could you explain the line underneath that one, the one starting with (n 0)? I think they subbed in k = 0 to get the first term, but I’m not sure about the other two terms. Thank you again, i really appreciate it

Reply 9

mqb2766

21

Original post

by loik24

Ohhh thank you, that’s okay. Then could you explain the line underneath that one, the one starting with (n 0)? I think they subbed in k = 0 to get the first term, but I’m not sure about the other two terms. Thank you again, i really appreciate it

Pretty much the whole point of this derivation is you want to do the next line of pascals triangle, so nCk, so you want to add the two existing nCk's, suitably indexed, together to get a new (n+1)Ck, so the next line / induction step. In pascals triangle, you can add the two elements above the new one apart from the first and last one which are the nC0 and nCn ones which lie outside the summation in the above proof.

Tbh, its just a glorified version of the binomial expansion so (f + g)^n where the powers are equivalent to the derivatives and the coefficients are nCk, as you should know. Then work through the induction for (f + g)^(n+1) and you need to make equivalent arguments, so add two nCk's together to get the new (n+1)Ck.

To look beyond the notation, just write down a few lines of pascais triangle / leibnitz and work out what the steps mean so each line is an extra derivative
fg
f'g + fg'
f''g + 2f'g' + fg''
f'''g + 3f''g' + 3f'g'' + fg'''
....
the posted proof in the op just generalises this / obfuscates with notation.

(edited 9 months ago)

Help with Edexcel Further Pure challenge question

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