please help - two questions (integration, induction) (1 Viewer)

[Lexie1001]

ok this induction q i just have no idea how to do...i got to the n=k+1 bit and it was impossible from there.
1. Use mathematical induction to prove n² > 10n + 7 for all n>= 11.
I'll type out my working later if anyone is desperate to see it.

2. Find S x(1-x²)⁴ using the substitution u=1-x.
I just dont know where to start with this one...how do you use the substitution?

S=integration symbol thing

Thanks guys

 

[Mountain.Dew]

ok this induction q i just have no idea how to do...i got to the n=k+1 bit and it was impossible from there.
1. Use mathematical induction to prove n² > 10n + 7 for all n>= 11.
I'll type out my working later if anyone is desperate to see it.

2. Find S x(1-x²)⁴ using the substitution u=1-x.
I just dont know where to start with this one...how do you use the substitution?

S=integration symbol thing

Thanks guys

i'll do q1

Use mathematical induction to prove n² > 10n + 7 for all n>= 11.

now i assume we have the assumption that k² > 10k + 7

so, RTP: (k+1)² > 10(k+1) + 7

now, LHS = (k+1)²
=k² + 2k + 1
> 10k + 7 + 2k + 1 = 12k + 8

now, IS 12k + 8 > 10(k+1) + 7?

we have: 12k + 8 > 10k + 10 + 7
2k > 9
k > 9/2 ==> true, since k > = 11

therefore, 12k + 8 > 10(k+1) + 7

therefore, (k+1)² > 10(k+1) + 7

and the additional statement at the end as well.

hopefully this is a logically sound method, im sure there are other ways to do it.

 

[Lexie1001]

^^^ i wasnt sure how to use the assumption, i think thats where i went wrong. thanks heaps for that!

 

[Mountain.Dew]

^^^ i wasnt sure how to use the assumption, i think thats where i went wrong. thanks heaps for that!

no problems Lexie1001!

with q2, i have fiddled around with the numbers a bit, and i dont seem to get anywhere better. perhaps i havent found the 'right' manipulation yet.

in the meantime, i would use the binomial theorem to expand the whole thing and integrate all of the terms individually. that is the LONG, inefficient, tediousBUT proven, "Tired and True" method.

 

[insert-username]

2. Find S x(1-x²)⁴ using the substitution u=1-x.
I just dont know where to start with this one...how do you use the substitution?

S=integration symbol thing

Thanks guys

Are you sure the substitution's u = (1-x)? That makes life very difficult, if not impossible. Using the substitution u = 1-x² is straightforward (find du/dx and hence du, sustitute, integrate, substitute original expression back in).


I_F

 

[Mountain.Dew]

Are you sure the substitution's u = (1-x)? That makes life very difficult, if not impossible. Using the substitution u = 1-x² is straightforward (find du/dx and hence du, sustitute, integrate, substitute original expression back in).
I_F

yesh, that is a much much better substitution. MUCH BETTER.

 

[Lexie1001]

hmmm ok ill try using u=(1-x)² my teacher probs made a mistake typying it out grrr.

 

[insert-username]

hmmm ok ill try using u=(1-x)² my teacher probs made a mistake typying it out grrr.

No, it's u = (1-x²). u = (1-x)² is just as hard as u = (1-x) to do.


I_F

 

[Lexie1001]

^^^ oh ok i think thats what i meant obviously i cant type either!