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Given the function \(y=x e^{2 x}\), use mathematical induction to prove that \(\dfrac{d^n y}{d x^n}=\left(2^n x+n 2^{n-1}\right) e^{2 x}\) for all positive integers \(n\), where \(\dfrac{d^n y}{d x^n}\) is the
\(n\)th derivative of \(y\) and \(\dfrac{d}{d x}\left(\dfrac{d^n y}{d x^n}\right)=\dfrac{d^{n+1} y}{d x^{n+1}}\). (3 marks)
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\(\text{Proof (See worked solutions)}\)
\(\text{Prove} \ \ \dfrac{d^n y}{d x^n}=\left(2^n x+n 2^{n-1}\right) e^{2 x}\)
\(\text{If } \ \ n=1:\)
\(\operatorname{LHS}=\dfrac{d}{d x}\left(x e^{2 x}\right)=x \cdot 2 e^{2 x}+1 \cdot e^{2 x}=(2 x+1) e^{2 x}\)
\(\operatorname{RHS}=\left(2^{1} \cdot x+1 \cdot 2^{0}\right) e^{2 x}=(2 x+1) e^{2 x}=\operatorname{RHS}\)
\(\therefore \ \text{True for} \ \ n=1\)
\(\text{Assume true for}\ \ n=k:\)
\(\dfrac{d^k y}{d x^k}=\left(2^k x+k\, 2^{k-1}\right) e^{2 x}\)
\(\text{Prove true for}\ \ n=k+1:\)
\(\text{i.e.}\ \dfrac{d^{k+1} y}{d x^{k+1}}=\left(2^{k+1} x+(k+1) 2^k\right) e^{2 x}\)
| \(\dfrac{d^{k+1} y}{d x^{k+1}}\) | \(=\dfrac{d}{d x}\left(2^k x+k\cdot 2^{k-1}\right) e^{2 x}\) |
| \(=\dfrac{d}{d x}\left(2^k x e^{2 x}+k\cdot 2^{k-1} e^{2 x}\right)\) | |
| \(=2^k x \cdot 2 e^{2 x}+2^k \cdot e^{2 x}+k \cdot 2^{k-1} \cdot 2 e^{2 x}\) | |
| \(=2^{k+1} x e^{2 x}+2^k \cdot e^{2 x}+k \cdot 2^k \cdot e^{2 x}\) | |
| \(=2^{k+1} x e^{2 x}+(k+1) 2^k \cdot e^{2 x}\) | |
| \(=\left(2^{k+1} x+(k+1) 2^k\right) e^{2 x}\) |
\(\Rightarrow \ \text{True for} \ \ n=k+1\)
\(\therefore \ \text{Since true for \(\ \ n=1\), by PMI, true for integers} \ \ n \geqslant 1.\)
A function \(f\) has the rule \(f(x)=x\,e^{2x}\).
Use mathematical induction to prove that \(f^{(n)}(x)=\big{(}2^{n}x+n\,2^{n-1}\big{)}e^{2x}\) for \(n \in \mathbb{Z}^{+}\), where \(f^{(n)}(x)\) represents the \(n\)th derivative of \(f(x)\). That is, \(f(x)\) has been differentiated \(n\) times. (3 marks)
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\(\text{Proof (See Worked Solutions)}\)
\(\text{If}\ \ n=1:\)
\(f(x)=x\,e^{2x}\ \ \Rightarrow \ \ f^{′}(x)=e^{2x} + 2x\,e^{2x} \)
\(f^{(1)}(x)=\big{(}2^{1}x + 1 \times 2^{1-1}\big{)}e^{2x} = e^{2x} + 2x\,e^{2x} \)
\(\therefore\ \text{True for}\ \ n=1.\)
\(\text{Assume true for}\ \ n=k \)
\(f^{(k)}(x)=\big{(}2^{k}x + k\,2^{k-1}\big{)}e^{2x}\)
\(\text{Prove true for}\ \ n=k+1 \)
\(\text{i.e.}\ f^{(k+1)}(x)=\big{(}2^{k+1}x + (k+1)2^{k}\big{)}e^{2x}\)
| \(\text{LHS}\) | \(= \dfrac{d}{dx} \big{(}f^{(k)}(x) \big{)} \) | |
| \(= \dfrac{d}{dx} \Big{[}\big{(}2^{k}x + k\,2^{k-1}\big{)}e^{2x}\Big{]} \) | ||
| \(= 2^{k}e^{2x}+(2^{k}x+k\,2^{k-1}) \times 2 e^{2x} \) | ||
| \(=e^{2x} \big{(}2^{k}+2\cdot2^{k}x + 2k \cdot 2^{k-1}\big{)}\) | ||
| \(=e^{2x}\big{(}2^{k+1}+2^{k}+k \cdot 2^{k} \big{)} \) | ||
| \(=\big{(}2^{k+1}x + (k+1)2^{k}\big{)}e^{2x}\) | ||
| \(=\ \text{RHS}\) |
\(\Rightarrow \text{True for}\ \ n=k+1\)
\(\therefore\ \text{Since true for}\ \ n=1\text{, by PMI, true for}\ n \in \mathbb{Z}^{+} \)
Using mathematical induction and integration by parts, show
`int_0^pi sin^nx\ dx=(n-1)/n int_0^pi sin^(n-2)x\ dx` for `n>=2.` (4 marks)
`text{Prove}\ \ int_0^pi sin^nx\ dx=(n-1)/n int_0^pi sin^(n-2)x\ dx\ \ text{for}\ \ n>=2.`
`text{If}\ \ n=2:`
| `text{LHS}` | `=int_0^pi sin^(2)x\ dx` | |
| `=1/2 int_0^pi (1-cos (2x))\ dx` | ||
| `=1/2 [x-1/2sin(2x)]_0^pi` | ||
| `=1/2[(pi-1/2sin(2pi))-0]` | ||
| `=pi/2` |
| `text{RHS}` | `=(2-1)/2 int_0^pi sin^0 x\ dx` | |
| `=1/2int_0^pi 1\ dx` | ||
| `=1/2[x]_0^pi` | ||
| `=pi/2` | ||
| `=\ text{LHS}` |
`:.\ text{True for}\ \ n=2`
`text{Assume true for}\ \ n=k`
`text{i.e.}\ int_0^pi sin^kx\ dx=(k-1)/k int_0^pi sin^(k-2)x\ dx`
`text{Prove true for}\ \ n=k+1`
`text{i.e.}\ int_0^pi sin^(k+1)x\ dx=(k)/(k+1) int_0^pi sin^(k-1)x\ dx`
`text{Let}\ \ I=\ text{LHS} = int_0^pi sin^(k+1)x\ dx=int_0^pi sinx * sin^(k)x\ dx`
`text{Using IBP:}`
| `u` | `=sin^k x` | `v^(′)=sinx` |
| `u^(′)` | `=kcosx*sin^(k-1)x` | `v=-cosx` |
| `I` | `=[uv]_0^pi-int_0^pi u^(′)v\ dx` | |
| `=[-sin^kx*cosx]_0^pi-int_0^pi -kcosx*sin^(k-1)x*cosx\ dx` | ||
| `=[(-sin^k pi*cos pi)-(-sin^k 0*cos 0)+kint cos^2x*sin^(k-1)x\ dx` | ||
| `=0+kint (1-sin^2x)*sin^(k-1)x\ dx` | ||
| `=kint_0^pi sin^(k-1)x\ dx-kint_0^pi sin^(k+1)x\ dx` | ||
| `I` | `=kint_0^pi sin^(k-1)x\ dx-kI` |
| `I+kI` | `=kint_0^pi sin^(k-1)x\ dx` | |
| `I(k+1)` | `=kint_0^pi sin^(k-1)x\ dx` | |
| `I` | `=k/(k+1) int_0^pi sin^(k-1)x\ dx` | |
| `=\ text{RHS}` |
`:.\ text{True for}\ \ n=k+1`
`:.\ text{Since true for} \ n=2,\ text{by PMI, true for integers} \ n>=2.`
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i. `text(Prove)\ \ d/(dx)(x)=1\ \ text(from first principles.)`
| `d/(dx)(x)` | `=lim_(h->0) (f(x+h)-f(x))/h` |
| `=lim_(h->0)(x+h-x)/h` | |
| `=1\ \ \ text(… as required)` |
ii. `text(Prove)\ \ d/(dx)(x^n)=nx^(n-1)\ \ text(for integers)\ n>=1`
`text(If)\ n=1,`
`text(LHS)=d/(dx)(x^1)=1`
`text(RHS)=(1)x^0=1=text(LHS)`
`:.\ text(True for)\ n=1`
`text(Assume true for)\ n=k`
`text(i.e.)\ \ d/(dx)(x^k)=kx^(k-1)`
`text(Prove true for)\ n=k+1`
`text(i.e.)\ \ d/(dx)x^(k+1)=(k+1)x^k`
| `text(LHS)` | `=d/(dx)(x^(k+1))` |
| `=d/(dx) (x*x^k)` | |
| `=d/(dx)(x) * x^k+x *d/(dx)(x^k)` | |
| `=(1xx x^k)+(x xxkx^(k-1))` | |
| `=x^k+kx^k` | |
| `=(k+1)x^k` | |
| `=\ text(RHS … as required)` |
`=>\ text(True for)\ n=k+1`
`:.\ text(S)text(ince it is true for)\ n=1, text(by PMI, true for integral)\ n>=1`.