/sci/ - Science & Math

[OP No.4890372]

Lets see if sci actually knows math.

[Anonymous No.4890374]

the answer is 7 prove me wrong.

[Anonymous No.4890376]

You've totally rustled my jimmies, I will now reclaim them by proving I can do your homework.

[Anonymous No.4890381]

>>4890372
> AB Calc
> real math

[Anonymous No.4890384]

>>4890381
Numerical Analysis, want a hint?

[Anonymous No.4890385]

>>4890374

I'm still not proven wrong... I think op wants homework help

[Anonymous No.4890386]

Sorry, not gonna do your undergrad physics homework for you.

[Anonymous No.4890391]

>>4890385

come on op I'm in high-school and got the answer

[Anonymous No.4890428]

>>4890372
thread smells of CS majors

l_i(x) is zero at x_n≠x_i and one at x_i
Clearly l*(x)=w(x)/(x-x_i) satisfies the zero property
l*(x_i)=∏(x_n-x_i) with n≠i
w'(x_i)=<span class="math">∑_{j=0}^{j≤n}∏_{k=0}^{k≤n} (x_i-x_k≠j)<span class="math">
in every term of the sum with j≠i there exist (x_i-x_i)=0 in the product leaving only j==i
∏_{k=0}^{k≤n} (x_i-x_k≠i)==l*(x_i)
clearly l*(x)/l*(x_i) satisfy the one property so l(x)=l*(x)/l*(x_i)=w(x)/w'(x)(x-x_i) ∎[/spoiler][/spoiler]

[Anonymous No.4890435]

>>4890372 (OP)
thread smells of CS majors

<span class="math">l_i(x)[/spoiler] is zero at <span class="math">x_n≠x_i[/spoiler] and one at <span class="math">x_i[/spoiler]
Clearly <span class="math">l*(x)=w(x)/(x-x_i)[/spoiler] satisfies the zero property
<span class="math">l*(x_i)=∏(x_n-x_i)[/spoiler] with n≠i

<span class="math">∑_{j=0}^{j≤n}∏_{k=0}^{k≤n} (x_i-x_k≠j)[/spoiler] in every term of the sum with j≠i there exist <span class="math">(x_i-x_i)=0[/spoiler] in the product leaving only <span class="math">j==i ∏_{k=0}^{k≤n} (x_i-x_k≠i)==l*(x_i)[/spoiler] hence <span class="math">w'(x)=l*(x_i)[/spoiler] and clearly <span class="math">l*(x)/l*(x_i)[/spoiler] satisfy the both properties so <span class="math">l(x)=l*(x)/l*(x_i)=w(x)/w'(x)(x-x_i)[/spoiler] ∎

[Anonymous No.4890443]

>>4890372
thread smells of CS majors

<span class="math">l_i(x)[/spoiler] is zero at <span class="math">x_n≠x_i[/spoiler] and one at <span class="math"> x_i[/spoiler]
Clearly <span class="math">l^*(x)=w(x)/(x-x_i)[/spoiler] satisfies the zero property
<span class="math">l^*(x_i)=∏_{n≠i}(x_n-x_i)[/spoiler]

∑_{j=0}^{j≤n}∏_{k=0}^{k≤n} (x_i-x_k≠j) in every term of the sum with j≠i there exist (x_i-x_i)=0 in the product leaving only j==i <span class="math">∏_{k=0}^{k≤n} (x_i-x_k≠i)==l^*(x_i)[/spoiler] hence <span class="math">w'(x)=l^{*}(x_i)[/spoiler] and clearly <span class="math">l^*(x)/l^*(x_i)[/spoiler] satisfy both property so <span class="math">l(x)=l^*(x)/l^*(x_i)=w(x)/w'(x)(x-x_i)[/spoiler] ∎

[Anonymous No.4890445]

>>4890372
thread smells of CS majors

<span class="math">l_i(x)[/spoiler] is zero at <span class="math">x_n≠x_i[/spoiler] and one at <span class="math"> x_i[/spoiler]
Clearly <span class="math">l^*(x)=w(x)/(x-x_i)[/spoiler] satisfies the zero property
<span class="math">l^*(x_i)=∏_{n≠i}(x_n-x_i)[/spoiler]

<span class="math">∑_{j=0}^{j≤n}∏_{k=0}^{k≤n} (x_i-x_k≠j)[/spoiler] in every term of the sum with j≠i there exist (x_i-x_i)=0 in the product leaving only j==i <span class="math">∏_{k=0}^{k≤n} (x_i-x_k≠i)==l^*(x_i)[/spoiler] hence <span class="math">w(x)=l^{*}(x_i)[/spoiler] and clearly <span class="math">l^*(x)/l^*(x_i)[/spoiler] satisfy the one property so <span class="math">l(x)=l^*(x)/l^*(x_i)=w(x)/w'(x)(x-x_i)[/spoiler] ∎

[Anonymous No.4890448]

>>4890445
If being a CS major means you don't help idiots with their homework, I'll gladly take them over you.

[Anonymous No.4890459]

trivial

[Anonymous No.4890485]

Is this really what CS majors do?

[Anonymous No.4890494]

>>4890485
>try to do
yes

[Anonymous No.4890528]

Is this really what CS majors solve?

[Anonymous No.4890545]

Get your shit together if you cannot solve this. Actually there is nothing to solve, really.

[Anonymous No.4890569]

>>4890374

yet to be proven wrong

[Anonymous No.4890586]

>>4890372
So in order for someone to prove to you he can math , is to write down some proof.
You can't even coem up with your own set of problems for us to solve? How pathetic are you? You are an utter disgrace to /sci/.